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1582. Special Positions in a Binary Matrix.md

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1582. Special Positions in a Binary Matrix

the first one in Weekly Contest 206.

Difficulty : Easy

Related Topics : Array

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

  • mine
    • Java
      • Runtime: 3 ms, faster than 25.00%, Memory Usage: 49.7 MB, less than 25.00% of Java online submissions 
        //O(R(R + 2C))time
//O(1)space
public int numSpecial(int[][] mat) {
    int r = mat.length;
    int c = mat[0].length;
    int res = 0;
    for(int i = 0; i < r; i++){
        for(int j = 0; j < c; j++){
            if(mat[i][j] == 1){
                if(check(mat, i, j)) res++;

                break;
            }
        }
    }
    return res;
}

boolean check(int[][] mat, int i, int j){
    int count = 0;
    for(int a = 0; a < mat[i].length; a++){
        if(mat[i][a] == 1) count++;
    }
    for(int a = 0; a < mat.length; a++){
        if(mat[a][j] == 1) count++;
    }
    return count == 2;
}
  • the most votes
  • Runtime: 3 ms, faster than 25.00%, Memory Usage: 39.4 MB, less than 100.00% of Java online submissions 
    // O(R * C)time
// O(R + C)space
public int numSpecial(int[][] mat) {
    int r = mat.length, c = mat[0].length;
    int[] row = new int[r], col = new int[c];
    for(int i = 0; i < r; i++) {
        for(int j = 0; j < c; j++) {
            if(mat[i][j] == 1) {
                row[i]++;
                col[j]++;
            }
        }
    }

    int count = 0;
    for(int i = 0; i < r; i++)
        for(int j = 0; j < c; j++)
            if(mat[i][j] == 1 && row[i] == 1 && col[j] == 1)
                count++;

    return count;
}