Sync LeetCode submission Runtime - 70 ms (5.65%), Memory - 14.8 MB (100.00%)

Author: jimit105

0074-search-a-2d-matrix/README.md

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+<p>You are given an <code>m x n</code> integer matrix <code>matrix</code> with the following two properties:</p>
+
+<ul>
+	<li>Each row is sorted in non-decreasing order.</li>
+	<li>The first integer of each row is greater than the last integer of the previous row.</li>
+</ul>
+
+<p>Given an integer <code>target</code>, return <code>true</code> <em>if</em> <code>target</code> <em>is in</em> <code>matrix</code> <em>or</em> <code>false</code> <em>otherwise</em>.</p>
+
+<p>You must write a solution in <code>O(log(m * n))</code> time complexity.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/10/05/mat.jpg" style="width: 322px; height: 242px;" />
+<pre>
+<strong>Input:</strong> matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
+<strong>Output:</strong> true
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/10/05/mat2.jpg" style="width: 322px; height: 242px;" />
+<pre>
+<strong>Input:</strong> matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
+<strong>Output:</strong> false
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>m == matrix.length</code></li>
+	<li><code>n == matrix[i].length</code></li>
+	<li><code>1 &lt;= m, n &lt;= 100</code></li>
+	<li><code>-10<sup>4</sup> &lt;= matrix[i][j], target &lt;= 10<sup>4</sup></code></li>
+</ul>

0074-search-a-2d-matrix/solution.py

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+# Approach 1 - Binary Search
+
+# Time: O(log(mn)), Space: O(1)
+
+class Solution:
+    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+        if not matrix:
+            return False
+        
+        m = len(matrix)
+        n = len(matrix[0])
+        
+        left, right = 0, m * n - 1
+        
+        while left <= right:
+            pivot_idx = left + (right - left) // 2
+            pivot_element = matrix[pivot_idx // n][pivot_idx % n]
+            
+            if target == pivot_element:
+                return True
+            elif target < pivot_element:
+                right = pivot_idx - 1
+            else:
+                left = pivot_idx + 1
+                
+        return False