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

/**
 * @param {number[][]} mat
 * @return {number}
 */
var numSpecial = function(mat) {
    var res = 0;
    for (var i = 0; i < mat.length; i++) {
        var specialCol = -1;
        var hasOnlyOne = false;
        for (var j = 0; j < mat[i].length; j++) {
            if (mat[i][j] === 1) {
                if (specialCol === -1) {
                    specialCol = j;
                    hasOnlyOne = true;
                } else {
                    hasOnlyOne = false;
                }
            }
        }
        if (!hasOnlyOne) continue;
        var isValid = true;
        for (var k = 0; k < mat.length; k++) {
            if (mat[k][specialCol] === 1 && k !== i) {
                isValid = false;
                break;
            }
        }
        if (isValid) res += 1;
    }
    return res;
};

Explain:

nope.

Complexity:

• Time complexity : O(n * m).
• Space complexity : O(1).