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Special Positions in a Binary Matrix - Array - Easy - LeetCode - MiniTV

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][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)