LeetCode in Kotlin

994. Rotting Oranges

Medium

You are given an m x n grid where each cell can have one of three values:

• 0 representing an empty cell,
• 1 representing a fresh orange, or
• 2 representing a rotten orange.

Every minute, any fresh orange that is 4-directionally adjacent to a rotten orange becomes rotten.

Return the minimum number of minutes that must elapse until no cell has a fresh orange. If this is impossible, return -1.

Example 1:

Input: grid = [[2,1,1],[1,1,0],[0,1,1]]

Output: 4

Example 2:

Input: grid = [[2,1,1],[0,1,1],[1,0,1]]

Output: -1

Explanation: The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.

Example 3:

Input: grid = [[0,2]]

Output: 0

Explanation: Since there are already no fresh oranges at minute 0, the answer is just 0.

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 10
• grid[i][j] is 0, 1, or 2.

Solution

class Solution {
    fun orangesRotting(grid: Array<IntArray>): Int {
        val queue = ArrayDeque<IntArray>()
        val row = grid.size
        val col: Int = grid[0].size
        var countActive = 0
        for (i in 0 until row) {
            for (j in 0 until col) {
                if (grid[i][j] == 2) {
                    queue.add(intArrayOf(i, j))
                }
                if (grid[i][j] != 0) {
                    countActive++
                }
            }
        }
        if (countActive == 0) {
            return 0
        }
        var countCurrent = 0
        var count = 0
        val dx = intArrayOf(0, 0, 1, -1)
        val dy = intArrayOf(1, -1, 0, 0)
        while (queue.isNotEmpty()) {
            val size: Int = queue.size
            count += size
            for (i in 0 until size) {
                val arr: IntArray = queue.removeFirst()
                for (j in 0..3) {
                    val x = arr[0] + dx[j]
                    val y = arr[1] + dy[j]
                    if (x < 0 || y < 0 || x >= row || y >= col || grid[x][y] != 1) {
                        continue
                    }
                    grid[x][y] = 2
                    queue.add(intArrayOf(x, y))
                }
            }
            if (queue.isNotEmpty()) {
                countCurrent++
            }
        }
        return if (countActive == count) countCurrent else -1
    }
}

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