class Solution{
    public:
        int numSpecial(vector<vector<int>> &mat){
            int ans = 0;
            int m = mat.size();
            int n = mat[0].size();
            for(int row = 0;row<m;row++){
                int colIndex = -1;
                int count = 0;
                for(int col = 0;col<n;col++){
                    if(mat[row][col]==1){
                        count++;
                        colIndex = col;
                    }
                }
                if(count==1){
                bool good = true;
            for(int r=0;r<m;r++){

                if(r!=row && mat[r][colIndex]==1) {
                    good = false;
                    break;
                }
            }
            if(good) ans++;
            }
            }
            return ans;
        }
};

/*
Approach:

We iterate through each row of the matrix and count how many 1s appear in that row.
While scanning the row, we also record the column index where the 1 occurs. If a row
contains exactly one 1, it becomes a candidate for a special position. For that candidate,
we then check the entire column corresponding to that 1 to ensure no other row contains a 1
in the same column. If the column also has exactly one 1 (the current position), then that
position satisfies both conditions and we increment the answer. After checking all rows,
the total number of such positions is returned.

Time Complexity: O(mn + m²) in the worst case. We scan each row once (mn work), and for
rows that contain exactly one 1 we scan the column (up to m operations). In practice this
is usually closer to O(mn).

Space Complexity: O(1) 
*/