search-2d-matrix py3

0074_search-a-2d-matrix/python3/solution.py

@@ -0,0 +1,42 @@
+class Solution:
+    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+        '''
+        Two binary searches, first across rows and then
+        across columns to quickly find the target
+        '''
+        
+        mid = -1
+        start, end = 0, len(matrix) - 1
+        
+        # Figure out which row the number belongs to
+        while start <= end:
+            mid = (start + end) // 2
+            
+            if target > matrix[mid][-1]:
+                start = mid + 1
+            elif target < matrix[mid][0]:
+                end = mid - 1
+            else:
+                break
+                
+        # If we never managed to find a row, that means number
+        # doesn't exist
+        if not start <= end:
+            return False
+        
+        # The row to search
+        row = matrix[mid]
+        start, end = 0, len(row) - 1
+        
+        while start <= end:
+            mid = (start + end) // 2
+            
+            if target == row[mid]:
+                return True
+            elif target > row[mid]:
+                start = mid + 1
+            else:
+                end = mid - 1
+        
+        return False
+