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.
public class Solution {
public int NumSpecial(int[][] mat) {
int m = mat.Length;
int n = mat[0].Length;
int res = 0;
var col = new int[n];
var row = new int[m];
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if (mat[i][j] == 1){
col[j]++;
row[i]++;
}
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;
}
}Time Complexity: O(n)
Space Complexity: O(n)
