Count Negative Numbers in a Sorted Matrix

Given a m x 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

 

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 100
  • -100 <= grid[i][j] <= 100

 

Follow up: Could you find an O(n + m) solution?SOLUTION:
class Solution:
    def countNegatives(self, grid: List[List[int]]) -> int:
        m = len(grid)
        n = len(grid[0])
        ctr = 0
        for i in range(m):
            beg = 0
            end = n - 1
            while beg <= end:
                mid = (beg + end) // 2
                if beg == end:
                    if grid[i][beg] < 0:
                        ctr += (n - beg)
                    break
                elif grid[i][mid] >= 0:
                    beg = mid + 1
                else:
                    end = mid
        return ctr

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