LeetCode in Kotlin

1391. Check if There is a Valid Path in a Grid

Medium

You are given an m x n grid. Each cell of grid represents a street. The street of grid[i][j] can be:

• 1 which means a street connecting the left cell and the right cell.
• 2 which means a street connecting the upper cell and the lower cell.
• 3 which means a street connecting the left cell and the lower cell.
• 4 which means a street connecting the right cell and the lower cell.
• 5 which means a street connecting the left cell and the upper cell.
• 6 which means a street connecting the right cell and the upper cell.

You will initially start at the street of 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). The path should only follow the streets.

Notice that you are not allowed to change any street.

Return true if there is a valid path in the grid or false otherwise.

Example 1:

Input: grid = [[2,4,3],[6,5,2]]

Output: true

Explanation: As shown you can start at cell (0, 0) and visit all the cells of the grid to reach (m - 1, n - 1).

Example 2:

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

Output: false

Explanation: As shown you the street at cell (0, 0) is not connected with any street of any other cell and you will get stuck at cell (0, 0)

Example 3:

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

Output: false

Explanation: You will get stuck at cell (0, 1) and you cannot reach cell (0, 2).

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• 1 <= grid[i][j] <= 6

Solution

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

class Solution {
    private val dirs = arrayOf(
        arrayOf(intArrayOf(0, -1), intArrayOf(0, 1)),
        arrayOf(intArrayOf(-1, 0), intArrayOf(1, 0)),
        arrayOf(
            intArrayOf(0, -1),
            intArrayOf(1, 0)
        ),
        arrayOf(intArrayOf(0, 1), intArrayOf(1, 0)),
        arrayOf(intArrayOf(0, -1), intArrayOf(-1, 0)),
        arrayOf(
            intArrayOf(0, 1),
            intArrayOf(-1, 0)
        )
    )

    // the idea is you need to check port direction match, you can go to next cell and check whether
    // you can come back.
    fun hasValidPath(grid: Array<IntArray>): Boolean {
        val m = grid.size
        val n = grid[0].size
        val visited = Array(m) { BooleanArray(n) }
        val q: Queue<IntArray> = LinkedList()
        q.add(intArrayOf(0, 0))
        visited[0][0] = true
        while (q.isNotEmpty()) {
            val cur = q.poll()
            val x = cur[0]
            val y = cur[1]
            val num = grid[x][y] - 1
            for (dir in dirs[num]) {
                val nx = x + dir[0]
                val ny = y + dir[1]
                if (nx < 0 || nx >= m || ny < 0 || ny >= n || visited[nx][ny]) {
                    continue
                }
                // go to the next cell and come back to orign to see if port directions are same
                for (backDir in dirs[grid[nx][ny] - 1]) {
                    if (nx + backDir[0] == x && ny + backDir[1] == y) {
                        visited[nx][ny] = true
                        q.add(intArrayOf(nx, ny))
                    }
                }
            }
        }
        return visited[m - 1][n - 1]
    }
}

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