count-negative-numbers-in-a-sorted-matrix.cpp

//question//

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?


//solution//

class Solution {
public:
    int countNegatives(vector<vector<int>>& grid) {
        int count=0;
        int n = grid.size();
        int m = grid[0].size();
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(grid[i][j]<0)
                {
                    count++;
                }
            }
        }
        return count;
    }
};