LeetCode in Kotlin

840. Magic Squares In Grid

Medium

A 3 x 3 magic square is a 3 x 3 grid filled with distinct numbers from 1 to 9 such that each row, column, and both diagonals all have the same sum.

Given a row x col grid of integers, how many 3 x 3 “magic square” subgrids are there? (Each subgrid is contiguous).

Example 1:

Input: grid = [[4,3,8,4],[9,5,1,9],[2,7,6,2]]

Output: 1

Explanation: The following subgrid is a 3 x 3 magic square: while this one is not: In total, there is only one magic square inside the given grid.

Example 2:

Input: grid = [[8]]

Output: 0

Constraints:

• row == grid.length
• col == grid[i].length
• 1 <= row, col <= 10
• 0 <= grid[i][j] <= 15

Solution

class Solution {
    fun numMagicSquaresInside(grid: Array<IntArray>): Int {
        val m = grid.size
        val n = grid[0].size
        var count = 0
        for (i in 0 until m - 2) {
            for (j in 0 until n - 2) {
                val set: MutableSet<Int> = HashSet()
                val sum = grid[i][j] + grid[i][j + 1] + grid[i][j + 2]
                if (sum == grid[i + 1][j] + grid[i + 1][j + 1] + grid[i + 1][j + 2] && sum == grid[i + 2][j] +
                    grid[i + 2][j + 1] + grid[i + 2][j + 2] && sum == grid[i][j] + grid[i + 1][j] + grid[i + 2][j] &&
                    sum == grid[i][j + 1] + grid[i + 1][j + 1] + grid[i + 2][j + 1] && sum == grid[i][j + 2] +
                    grid[i + 1][j + 2] + grid[i + 2][j + 2] && sum == grid[i][j] + grid[i + 1][j + 1] +
                    grid[i + 2][j + 2] && sum == grid[i][j + 2] + grid[i + 1][j + 1] + grid[i + 2][j] && set.add(
                        grid[i][j],
                    ) &&
                    isLegit(grid[i][j]) &&
                    set.add(grid[i][j + 1]) &&
                    isLegit(grid[i][j + 1]) &&
                    set.add(grid[i][j + 2]) &&
                    isLegit(grid[i][j + 2]) &&
                    set.add(grid[i + 1][j]) &&
                    isLegit(grid[i + 1][j]) &&
                    set.add(grid[i + 1][j + 1]) &&
                    isLegit(grid[i + 1][j + 1]) &&
                    set.add(grid[i + 1][j + 2]) &&
                    isLegit(grid[i + 1][j + 2]) &&
                    set.add(grid[i + 2][j]) &&
                    isLegit(grid[i + 2][j]) &&
                    set.add(grid[i + 2][j + 1]) &&
                    isLegit(grid[i + 2][j + 1]) &&
                    set.add(grid[i + 2][j + 2]) &&
                    isLegit(grid[i + 2][j + 2])
                ) {
                    count++
                }
            }
        }
        return count
    }

    private fun isLegit(num: Int): Boolean {
        return num in 1..9
    }
}

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