Count Negatives in a Sorted Matrix

/*
 * Problem : Process a matrix.
 * Input   : r and c, then r rows of c integers.
 * Output  : The count of negatives.
 * Approach: A sorted matrix.
 * Time    : O(r + c)
 * Space   : O(1)
 * Java    : 8 or newer.
 */
import java.util.Scanner;

public class Main {
    // This method solves the task. It receives the input and returns the answer.
    static long solve(int[][] m) {
        // Step 1: Start at the bottom-left cell. Moving right makes numbers smaller, moving up makes them bigger.
        int row = m.length - 1;
        int col = 0;
        long count = 0;
        while (row >= 0 && col < m[0].length) {
            if (m[row][col] < 0) {
                // Step 2: This cell is negative, so all cells to its right in this row are negative too. Count them and go up.
                count += m[0].length - col;
                row--;
            } else {
                // Step 3: This cell is not negative, so move right to find where the negatives start.
                col++;
            }
        }
        return count;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int rows = scanner.nextInt();
        int cols = scanner.nextInt();
        int[][] m = new int[rows][cols];
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                m[i][j] = scanner.nextInt();
            }
        }
        System.out.println(solve(m));
    }
}