From 50bf1933bbbcc29414dc6a8817095ee21b3857ce Mon Sep 17 00:00:00 2001
From: Ainur Dzhaianbaeva <adjayanbaeva@gmail.com>
Date: Sat, 25 Jun 2022 15:03:18 -0700
Subject: [PATCH 1/3] solve grouped anagrams problem

---
 hash_practice/exercises.py | 17 ++++++++++++++++-
 1 file changed, 16 insertions(+), 1 deletion(-)

diff --git a/hash_practice/exercises.py b/hash_practice/exercises.py
index 48bf95e..e4bbec2 100644
--- a/hash_practice/exercises.py
+++ b/hash_practice/exercises.py
@@ -5,7 +5,22 @@ def grouped_anagrams(strings):
         Time Complexity: ?
         Space Complexity: ?
     """
-    pass
+    if not strings:
+        return []
+
+    result = {}
+   
+    for word in strings:
+        key = ''.join(sorted(word))
+        if key in result:
+            result.get(key).append(word)
+        else:
+            result[key] = [word]
+    return result.values()
+    
+
+ 
+
 
 def top_k_frequent_elements(nums, k):
     """ This method will return the k most common elements

From 67c9bf32aa4472bb673673e10917057716c69ae8 Mon Sep 17 00:00:00 2001
From: Ainur Dzhaianbaeva <adjayanbaeva@gmail.com>
Date: Sat, 25 Jun 2022 16:43:36 -0700
Subject: [PATCH 2/3] solve top_k_frequent_elements problem

---
 hash_practice/exercises.py | 25 +++++++++++++++++++++----
 1 file changed, 21 insertions(+), 4 deletions(-)

diff --git a/hash_practice/exercises.py b/hash_practice/exercises.py
index e4bbec2..c9e951b 100644
--- a/hash_practice/exercises.py
+++ b/hash_practice/exercises.py
@@ -2,14 +2,13 @@
 def grouped_anagrams(strings):
     """ This method will return an array of arrays.
         Each subarray will have strings which are anagrams of each other
-        Time Complexity: ?
-        Space Complexity: ?
+        Time Complexity: O(nklog(k)), where k is the length of the string 
+        Space Complexity: O(log(n))
     """
     if not strings:
         return []
 
     result = {}
-   
     for word in strings:
         key = ''.join(sorted(word))
         if key in result:
@@ -28,7 +27,25 @@ def top_k_frequent_elements(nums, k):
         Time Complexity: ?
         Space Complexity: ?
     """
-    pass
+    if not nums:
+        return []
+    count = {}
+    frequency = [[] for i in range(len(nums)+1)]
+    for n in nums:
+        if n not in count:
+            count[n] = 0
+        count[n] += 1
+
+    for key, val in count.items():
+        frequency[val].append(key)
+    
+
+    result = []
+    for i in range(len(frequency)-1, 0, -1):
+        for n in frequency[i]:
+            result.append(n)
+            if len(result) == k:
+                return result
 
 
 def valid_sudoku(table):

From 4db1eeef78fead0183581f859b958d0077bee89a Mon Sep 17 00:00:00 2001
From: Ainur Dzhaianbaeva <adjayanbaeva@gmail.com>
Date: Sun, 17 Jul 2022 16:13:25 -0700
Subject: [PATCH 3/3] solve validSudoku() function

---
 hash_practice/exercises.py | 29 +++++++++++++++++++++++------
 1 file changed, 23 insertions(+), 6 deletions(-)

diff --git a/hash_practice/exercises.py b/hash_practice/exercises.py
index c9e951b..01ab9d9 100644
--- a/hash_practice/exercises.py
+++ b/hash_practice/exercises.py
@@ -24,8 +24,8 @@ def grouped_anagrams(strings):
 def top_k_frequent_elements(nums, k):
     """ This method will return the k most common elements
         In the case of a tie it will select the first occuring element.
-        Time Complexity: ?
-        Space Complexity: ?
+        Time Complexity: O(n)
+        Space Complexity: O(n)
     """
     if not nums:
         return []
@@ -54,8 +54,25 @@ def valid_sudoku(table):
         Each element can either be a ".", or a digit 1-9
         The same digit cannot appear twice or more in the same 
         row, column or 3x3 subgrid
-        Time Complexity: ?
-        Space Complexity: ?
-    """
-    pass
+        Time Complexity: o(n^2)
+        Space Complexity: O(n)
+#     """
+from collections import defaultdict     
+def validSudoku(board):
+    if board == None:
+        return None
+    N = 9
+    cols = defaultdict(set)
+    rows = defaultdict(set)
+    squares = defaultdict(set)
 
+    for row in range(9):
+        for col in range(9):
+            if board[row][col] == ".":
+                continue
+            if(board[row][col] in rows[row] or board[row][col] in cols[col] or board[row][col] in squares[(row // 3, col // 3)]):
+                return False
+            cols[col].add(board[row][col])
+            rows[row].add(board[row][col])
+            squares[(row // 3, col // 3)].add(board[row][col])
+    return True