[LeetCode Sync] Runtime - 1 ms (82.62%), Memory - 21.3 MB (51.24%)

Author: NinePiece2

Solutions/0378-kth-smallest-element-in-a-sorted-matrix/README.md

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+<p>Given an <code>n x n</code> <code>matrix</code> where each of the rows and columns is sorted in ascending order, return <em>the</em> <code>k<sup>th</sup></code> <em>smallest element in the matrix</em>.</p>
+
+<p>Note that it is the <code>k<sup>th</sup></code> smallest element <strong>in the sorted order</strong>, not the <code>k<sup>th</sup></code> <strong>distinct</strong> element.</p>
+
+<p>You must find a solution with a memory complexity better than <code>O(n<sup>2</sup>)</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8
+<strong>Output:</strong> 13
+<strong>Explanation:</strong> The elements in the matrix are [1,5,9,10,11,12,13,<u><strong>13</strong></u>,15], and the 8<sup>th</sup> smallest number is 13
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> matrix = [[-5]], k = 1
+<strong>Output:</strong> -5
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == matrix.length == matrix[i].length</code></li>
+	<li><code>1 &lt;= n &lt;= 300</code></li>
+	<li><code>-10<sup>9</sup> &lt;= matrix[i][j] &lt;= 10<sup>9</sup></code></li>
+	<li>All the rows and columns of <code>matrix</code> are <strong>guaranteed</strong> to be sorted in <strong>non-decreasing order</strong>.</li>
+	<li><code>1 &lt;= k &lt;= n<sup>2</sup></code></li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong>Follow up:</strong></p>
+
+<ul>
+	<li>Could you solve the problem with a constant memory (i.e., <code>O(1)</code> memory complexity)?</li>
+	<li>Could you solve the problem in <code>O(n)</code> time complexity? The solution may be too advanced for an interview but you may find reading <a href="http://www.cse.yorku.ca/~andy/pubs/X+Y.pdf" target="_blank">this paper</a> fun.</li>
+</ul>

Solutions/0378-kth-smallest-element-in-a-sorted-matrix/solution.py

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+class Solution:
+    def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
+        n = len(matrix)
+        left, right = matrix[0][0], matrix[-1][-1]
+
+        while right > left:
+            mid = (left + right) // 2
+            if self.checker(matrix, mid, k, n):
+                right = mid
+            else:
+                left = mid + 1
+        return left
+            
+    
+    def checker(self, matrix, mid, k, n):
+        count = 0
+        i, j = n - 1, 0
+        while i >= 0 and j < n:
+            if matrix[i][j] <= mid:
+                count += i + 1
+                j += 1
+            else:
+                i -= 1
+        return count >= k