LeetCode in Kotlin

1293. Shortest Path in a Grid with Obstacles Elimination

Hard

You are given an m x n integer matrix grid where each cell is either 0 (empty) or 1 (obstacle). You can move up, down, left, or right from and to an empty cell in one step.

Return the minimum number of steps to walk from the upper left corner (0, 0) to the lower right corner (m - 1, n - 1) given that you can eliminate at most k obstacles. If it is not possible to find such walk return -1.

Example 1:

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

Output: 6

Explanation:

The shortest path without eliminating any obstacle is 10.

The shortest path with one obstacle elimination at position (3,2) is 6. Such path is (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (3,2) -> (4,2).

Example 2:

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

Output: -1

Explanation: We need to eliminate at least two obstacles to find such a walk.

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 40
• 1 <= k <= m * n
• grid[i][j] is either 0 or 1.
• grid[0][0] == grid[m - 1][n - 1] == 0

Solution

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

class Solution {
    fun shortestPath(grid: Array<IntArray>, k: Int): Int {
        if (grid.size == 1 && grid[0].size == 1 && grid[0][0] == 0) {
            return 0
        }
        // 4 potential moves:
        val moves = arrayOf(intArrayOf(1, 0), intArrayOf(-1, 0), intArrayOf(0, 1), intArrayOf(0, -1))
        val m = grid.size
        val n = grid[0].size
        // use obs to record the min total obstacles when traverse to the position
        val obs = Array(m) { IntArray(n) }
        for (i in 0 until m) {
            for (j in 0 until n) {
                obs[i][j] = Int.MAX_VALUE
            }
        }
        obs[0][0] = 0
        // Queue to record {x cord, y cord, total obstacles when trvavers to this position}
        val que: Queue<IntArray> = LinkedList()
        que.add(intArrayOf(0, 0, 0))
        var level = 0
        while (que.isNotEmpty()) {
            val size = que.size
            level++
            for (i in 0 until size) {
                val current = que.poll()
                for (move in moves) {
                    val next = intArrayOf(current[0] + move[0], current[1] + move[1])
                    if (next[0] == m - 1 && next[1] == n - 1) {
                        return level
                    }
                    if (next[0] < 0 || next[0] > m - 1 || next[1] < 0 || next[1] > n - 1) {
                        continue
                    }
                    if (current[2] + grid[next[0]][next[1]] < obs[next[0]][next[1]] &&
                        current[2] + grid[next[0]][next[1]] <= k
                    ) {
                        obs[next[0]][next[1]] = current[2] + grid[next[0]][next[1]]
                        que.add(intArrayOf(next[0], next[1], obs[next[0]][next[1]]))
                    }
                }
            }
        }
        return -1
    }
}

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