1476 Count Negative Numbers In A Sorted Matrix

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?

class Solution:
    def countNegatives(self, grid: List[List[int]]) -> int:
        ans = 0
        for k in grid:
            left = 0
            right = len(k) - 1
            if k[right] >= 0:
                continue
            while(left < right):
                print(left, right)
                mid = (right + left) // 2
                if k[mid] < 0:
                    right = mid 
                else :
                    left = mid + 1
            ans += (len(k)   - right )
            #print(len(k)  - right)
        return ans