74. Search a 2D Matrix

Difficulty: Medium
136 / 136 test cases passed.
Runtime: 64 ms
Memory Usage: 15.9 MB

Author: YuriSpiridonov

Medium/74.Searcha2DMatrix.py

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+"""
+    Write an efficient algorithm that searches for a value in an m x n matrix. 
+    This matrix has the following properties:
+    - Integers in each row are sorted from left to right.
+    - The first integer of each row is greater than the last integer of the 
+      previous row.
+    
+    Example:
+    Input:
+    matrix = [
+      [1,   3,  5,  7],
+      [10, 11, 16, 20],
+      [23, 30, 34, 50]
+    ]
+    target = 3
+    Output: true
+"""
+#Difficulty: Medium
+#136 / 136 test cases passed.
+#Runtime: 64 ms
+#Memory Usage: 15.9 MB
+
+#Runtime: 64 ms, faster than 84.49% of Python3 online submissions for Search a 2D Matrix.
+#Memory Usage: 15.9 MB, less than 5.88% of Python3 online submissions for Search a 2D Matrix
+
+class Solution:
+    
+    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+        if not matrix or not matrix[0]: return
+        self.rows = len(matrix) - 1
+        self.cols = len(matrix[0]) - 1
+        return self.binarySearchRow(0, self.rows, matrix, target)
+        
+    def binarySearchRow(self, left, right, matrix, target):
+        while left <= right:
+            mid = (left + right) // 2
+            if matrix[mid][0] <= target <= matrix[mid][-1]:
+                return self.binarySearchTarget(0, self.cols, matrix[mid], target)
+            if matrix[mid][0] < target:
+                left = mid + 1
+            else:
+                right = mid - 1
+                
+    def binarySearchTarget(self, left, right, matrix, target):
+        while left <= right:
+            mid = (left + right) // 2
+            if matrix[mid] == target:
+                return True
+            if matrix[mid] < target:
+                left = mid + 1
+            else:
+                right = mid - 1
+        return False