leetcode 1582 - Special Positions in a Binary Matrix

https://leetcode.com/problems/special-positions-in-a-binary-matrix/

Given a rows x cols matrix mat, where mat[i][j] is either 0 or 1, return the number of special positions in mat.

A position (i,j) is called special if mat[i][j] == 1 and all other elements in row i and column j are 0 (rows and columns are 0-indexed).

Example 1:

Input: mat = [[1,0,0],
[0,0,1],
[1,0,0]]
Output: 1
Explanation: (1,2) is a special position because mat[1][2] == 1 and all other elements in row 1 and column 2 are 0.
Example 2:

Input: mat = [[1,0,0],
[0,1,0],
[0,0,1]]
Output: 3
Explanation: (0,0), (1,1) and (2,2) are special positions.
Example 3:

Input: mat = [[0,0,0,1],
[1,0,0,0],
[0,1,1,0],
[0,0,0,0]]
Output: 2
Example 4:

Input: mat = [[0,0,0,0,0],
[1,0,0,0,0],
[0,1,0,0,0],
[0,0,1,0,0],
[0,0,0,1,1]]
Output: 3

Constraints:

rows == mat.length
cols == mat[i].length
1 <= rows, cols <= 100
mat[i][j] is 0 or 1.

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class Solution {
public:
int numSpecial(vector<vector<int>>& mat) {
int r = mat.size();
int c = mat[0].size();

vector<int> rows(r, 0), cols(c, 0);

for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
if (mat[i][j] == 1) {
rows[i]++;
cols[j]++;
}
}
}

int total = 0;

for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
if (mat[i][j] == 1 && rows[i] == 1 && cols[j] == 1)
total++;
}
}

return total;
}
};