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.

countNegatives.js

/**
 * @param {number[][]} grid - A 2D matrix sorted in non-increasing order
 * @return {number} - Total count of negative numbers in the matrix
 */
var countNegatives = function (grid) {
    let count = 0;

    // Loop through each row
    for (let row = 0; row < grid.length; row++) {

        // Loop through each value in the current row
        for (let col = 0; col < grid[row].length; col++) {

            // If the value is negative, increment the counter
            if (grid[row][col] < 0) {
                count++;
            }
        }
    }

    return count;
};

countNegatives.py

class Solution(object):
    def countNegatives(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """

        # Number of rows (m) and columns (n)
        m = len(grid)
        n = len(grid[0])

        # Start at bottom-left corner
        row = m - 1
        col = 0

        count = 0

        # Walk while we stay inside the grid
        while row >= 0 and col < n:

            # If the current value is negative...
            if grid[row][col] < 0:
                # All values to the RIGHT are also negative because rows are sorted
                count += (n - col)

                # Move UP to check the next row
                row -= 1

            else:
                # Current value is non-negative, so move RIGHT to find negatives
                col += 1

        return count