1582. Special Positions in a Binary Matrix

Problem

Given an m x n binary matrix mat, 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.

  Constraints:

• m == mat.length

• n == mat[i].length

• 1 <= m, n <= 100

• mat[i][j] is either 0 or 1.

Solution (Java)

class Solution {
    public int numSpecial(int[][] mat) {
        int count = 0;
        for (int i = 0; i < mat.length; i++) {
            for (int j = 0; j < mat[0].length; j++) {
                if (mat[i][j] == 1 && isSpecial(mat, i, j)) {
                    count++;
                }
            }
        }
        return count;
    }

    private boolean isSpecial(int[][] mat, int row, int col) {
        for (int i = 0; i < mat.length; i++) {
            if (i != row && mat[i][col] == 1) {
                return false;
            }
        }
        for (int j = 0; j < mat[0].length; j++) {
            if (j != col && mat[row][j] == 1) {
                return false;
            }
        }
        return true;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).