Count Negative Numbers in a Sorted Matrix Program (Leetcode)

Count Negative Numbers in a Sorted Matrix Program (Leetcode):

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

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

 

Follow up: Could you find an O(n + m) solution?

class Solution:
    def binarySearch(self, nums):
        left = 0
        right = len(nums) - 1
        
        while left <= right:
            mid = (left + right) // 2
            if nums[mid] >= 0:
                left = mid + 1
            else:
                right = mid - 1
                
        return len(nums) - left
    
    def countNegatives(self, grid: List[List[int]]) -> int:
        count = 0
        for row in grid:
            count += self.binarySearch(row)
            
        return count