LeetCode in Kotlin

1368. Minimum Cost to Make at Least One Valid Path in a Grid

Hard

Given an m x n grid. Each cell of the grid has a sign pointing to the next cell you should visit if you are currently in this cell. The sign of grid[i][j] can be:

• 1 which means go to the cell to the right. (i.e go from grid[i][j] to grid[i][j + 1])
• 2 which means go to the cell to the left. (i.e go from grid[i][j] to grid[i][j - 1])
• 3 which means go to the lower cell. (i.e go from grid[i][j] to grid[i + 1][j])
• 4 which means go to the upper cell. (i.e go from grid[i][j] to grid[i - 1][j])

Notice that there could be some signs on the cells of the grid that point outside the grid.

You will initially start at the upper left cell (0, 0). A valid path in the grid is a path that starts from the upper left cell (0, 0) and ends at the bottom-right cell (m - 1, n - 1) following the signs on the grid. The valid path does not have to be the shortest.

You can modify the sign on a cell with cost = 1. You can modify the sign on a cell one time only.

Return the minimum cost to make the grid have at least one valid path.

Example 1:

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

Output: 3

Explanation: You will start at point (0, 0). The path to (3, 3) is as follows. (0, 0) –> (0, 1) –> (0, 2) –> (0, 3) change the arrow to down with cost = 1 –> (1, 3) –> (1, 2) –> (1, 1) –> (1, 0) change the arrow to down with cost = 1 –> (2, 0) –> (2, 1) –> (2, 2) –> (2, 3) change the arrow to down with cost = 1 –> (3, 3) The total cost = 3.

Example 2:

Input: grid = [[1,1,3],[3,2,2],[1,1,4]]

Output: 0

Explanation: You can follow the path from (0, 0) to (2, 2).

Example 3:

Input: grid = [[1,2],[4,3]]

Output: 1

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 100
• 1 <= grid[i][j] <= 4

Solution

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

@Suppress("NAME_SHADOWING")
class Solution {
    private val dir = arrayOf(
        intArrayOf(0, 0), intArrayOf(0, 1),
        intArrayOf(0, -1), intArrayOf(1, 0), intArrayOf(-1, 0)
    )

    fun minCost(grid: Array<IntArray>): Int {
        val visited = Array(grid.size) { IntArray(grid[0].size) }
        val queue: Queue<Pair> = LinkedList()
        addAllTheNodeInRange(0, 0, grid, queue, visited)
        if (visited[grid.size - 1][grid[0].size - 1] == 1) {
            return 0
        }
        var cost = 0
        while (queue.isNotEmpty()) {
            cost++
            val size = queue.size
            for (i in 0 until size) {
                val pa = queue.poll()
                for (k in 1 until dir.size) {
                    val m = Objects.requireNonNull(pa).x + dir[k][0]
                    val n = pa.y + dir[k][1]
                    addAllTheNodeInRange(m, n, grid, queue, visited)
                    if (visited[grid.size - 1][grid[0].size - 1] == 1) {
                        return cost
                    }
                }
            }
        }
        return -1
    }

    private fun addAllTheNodeInRange(
        x: Int,
        y: Int,
        grid: Array<IntArray>,
        queue: Queue<Pair>,
        visited: Array<IntArray>
    ) {
        var x = x
        var y = y
        while (x >= 0 && x < visited.size && y >= 0 && y < visited[0].size && visited[x][y] == 0) {
            queue.offer(Pair(x, y))
            visited[x][y]++
            val d = dir[grid[x][y]]
            x += d[0]
            y += d[1]
        }
    }

    private class Pair(var x: Int, var y: Int)
}

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