LeetCode in Kotlin

1582. Special Positions in a Binary Matrix

Easy

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

class Solution {
    fun numSpecial(mat: Array<IntArray>): Int {
        var count = 0
        for (i in mat.indices) {
            for (j in mat[0].indices) {
                if (mat[i][j] == 1 && isSpecial(mat, i, j)) {
                    count++
                }
            }
        }
        return count
    }

    private fun isSpecial(mat: Array<IntArray>, row: Int, col: Int): Boolean {
        for (i in mat.indices) {
            if (i != row && mat[i][col] == 1) {
                return false
            }
        }
        for (j in mat[0].indices) {
            if (j != col && mat[row][j] == 1) {
                return false
            }
        }
        return true
    }
}

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