binary serach -1 completed

Sample.py

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+# // Time Complexity : O(log(m*n))
+# // Space Complexity : O(1)
+# // Did this code successfully run on Leetcode : Yes
+# // Any problem you faced while coding this : No
+
+
+# // Your code here along with comments explaining your approach in three sentences only
+class Solution:
+    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
+        m = len(matrix)
+        n = len(matrix[0])
+        low = 0
+        high = (m*n) - 1
+        while low<=high:
+            mid = low + (high-low)//2
+
+            #convert index to position on matrix
+            r = mid//n 
+            c = mid%n
+
+            if matrix[r][c] == target:
+                return True
+            elif matrix[r][c] > target:
+                high = mid - 1
+            else:
+                low = mid + 1
+
+        return False