1351. Count Negative Numbers in a Sorted Matrix

Below is the top 7 difference between C++ Reference vs Pointer

Q

Given a m * 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

A:

class Solution {
public:
    int countNegatives(vector<vector<int>>& grid) {
        if ( grid.size()==0 || grid[0].size() == 0)
            return 0;
        return helper(grid,0,0,grid.size(),grid[0].size());
    }
private:
    int helper(vector<vector<int>>& grid, int X1, int Y1, const int m, const int n) {
        if (X1>=m ||  Y1>=n )
        {
            return 0;
        }
        if (grid[X1][Y1]< 0)
        {
            return (m-X1) * (n-Y1);
        }
        int res = 0;
        // linear search a row
        vector<int> & row = grid[X1];
        for (int j = Y1+1; j<n; ++j )
        {
            if(row[j]<0)
            {
                res += n-j;
                break;
            }
        }
            
        // linear search a column
        for (int i = X1+1; i<m; ++i)
        {
            if(grid[i][Y1]<0)
            {
                res += m -i;
                break;
            }
        }
        #
        res +=helper(grid, X1+1, Y1+1, m, n);
        return res;
    }
};