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.
Solution:
class Solution {
public int numSpecial(int[][] mat) {
int m = mat.length;
int n = mat[0].length;
int[] row = new int[m];
int[] col = new int[n];
for (int j = 0; j < n; j ++) {
for (int i = 0; i < m; i ++) {
row[i] += mat[i][j];
col[j] += mat[i][j];
}
}
int res = 0;
for (int i = 0; i < m; i ++) {
for (int j = 0; j < n; j ++) {
if (mat[i][j] == 1 && row[i] == 1 && col[j] == 1) {
res ++;
}
}
}
return res;
}
}