LeetCode in Kotlin

909. Snakes and Ladders

Medium

You are given an n x n integer matrix board where the cells are labeled from 1 to n2 in a Boustrophedon style starting from the bottom left of the board (i.e. board[n - 1][0]) and alternating direction each row.

You start on square 1 of the board. In each move, starting from square curr, do the following:

A board square on row r and column c has a snake or ladder if board[r][c] != -1. The destination of that snake or ladder is board[r][c]. Squares 1 and n2 do not have a snake or ladder.

Note that you only take a snake or ladder at most once per move. If the destination to a snake or ladder is the start of another snake or ladder, you do not follow the subsequent snake or ladder.

Return the least number of moves required to reach the square n2. If it is not possible to reach the square, return -1.

Example 1:

Input: board = [[-1,-1,-1,-1,-1,-1],[-1,-1,-1,-1,-1,-1],[-1,-1,-1,-1,-1,-1],[-1,35,-1,-1,13,-1],[-1,-1,-1,-1,-1,-1],[-1,15,-1,-1,-1,-1]]

Output: 4

Explanation:

In the beginning, you start at square 1 (at row 5, column 0).

You decide to move to square 2 and must take the ladder to square 15.

You then decide to move to square 17 and must take the snake to square 13.

You then decide to move to square 14 and must take the ladder to square 35.

You then decide to move to square 36, ending the game.

This is the lowest possible number of moves to reach the last square, so return 4.

Example 2:

Input: board = [[-1,-1],[-1,3]]

Output: 1

Constraints:

Solution

import java.util.LinkedList
import java.util.Queue

class Solution {
    private var size = 0
    fun snakesAndLadders(board: Array<IntArray>): Int {
        val queue: Queue<Int> = LinkedList()
        size = board.size
        val target = size * size
        val visited = BooleanArray(target)
        queue.add(1)
        visited[0] = true
        var step = 0
        while (queue.isNotEmpty()) {
            val queueSize = queue.size
            for (i in 0 until queueSize) {
                val previousLabel = queue.poll()
                if (previousLabel == target) {
                    return step
                }
                for (currentLabel in previousLabel + 1..Math.min(target, previousLabel + 6)) {
                    if (visited[currentLabel - 1]) {
                        continue
                    }
                    visited[currentLabel - 1] = true
                    val position = indexToPosition(currentLabel)
                    if (board[position[0]][position[1]] == -1) {
                        queue.add(currentLabel)
                    } else {
                        queue.add(board[position[0]][position[1]])
                    }
                }
            }
            step++
        }
        return -1
    }

    private fun indexToPosition(index: Int): IntArray {
        val vertical = size - 1 - (index - 1) / size
        val horizontal: Int = if ((size - vertical) % 2 == 1) {
            (index - 1) % size
        } else {
            size - 1 - (index - 1) % size
        }
        return intArrayOf(vertical, horizontal)
    }
}