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### by Sumit Chourasia | Sep 27, 2020 | Category :coding | Tags : algorithmarrayडेटा-संरचनाआसानleetcode #### Special Positions in a Binary Matrix - Array - Easy - LeetCode - मिनी टीवी

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 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.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)