Count Negative Numbers in a Sorted Matrix

Author: anantkaushik

1351-count-negative-numbers-in-a-sorted-matrix.py

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+"""
+Problem Link: https://leetcode.com/problems/count-negative-numbers-in-a-sorted-matrix/
+
+Given a m * n matrix grid which is sorted in non-increasing order both row-wise and column-wise. 
+
+Return the number of negative numbers in grid. 
+
+Example 1:
+Input: grid = [[4,3,2,-1],[3,2,1,-1],[1,1,-1,-2],[-1,-1,-2,-3]]
+Output: 8
+Explanation: There are 8 negatives number in the matrix.
+
+Example 2:
+Input: grid = [[3,2],[1,0]]
+Output: 0
+
+Example 3:
+Input: grid = [[1,-1],[-1,-1]]
+Output: 3
+
+Example 4:
+Input: grid = [[-1]]
+Output: 1
+ 
+Constraints:
+m == grid.length
+n == grid[i].length
+1 <= m, n <= 100
+-100 <= grid[i][j] <= 100
+"""
+# Time Complexity: O(m+n)
+# Space Complexity: O(1)
+class Solution:
+    def countNegatives(self, grid: List[List[int]]) -> int:
+        m, n = len(grid), len(grid[0])
+        row, column, count = m-1, 0, 0
+        while row >= 0 and column < n:
+          if grid[row][column] < 0:
+            count += (n-column)
+            row -= 1
+          else:
+            column += 1
+        return count
+        
+
+# Time Complexity: O(m*n)
+# Space Complexity: O(1)
+class Solution1:
+    def countNegatives(self, grid: List[List[int]]) -> int:
+        count = 0
+        for i in range(len(grid)):
+          for j in range(len(grid[0])):
+            if grid[i][j] < 0:
+              count += 1
+        return count