LeetCode in Kotlin

847. Shortest Path Visiting All Nodes

Hard

You have an undirected, connected graph of n nodes labeled from 0 to n - 1. You are given an array graph where graph[i] is a list of all the nodes connected with node i by an edge.

Return the length of the shortest path that visits every node. You may start and stop at any node, you may revisit nodes multiple times, and you may reuse edges.

Example 1:

Input: graph = [[1,2,3],[0],[0],[0]]

Output: 4

Explanation: One possible path is [1,0,2,0,3]

Example 2:

Input: graph = [[1],[0,2,4],[1,3,4],[2],[1,2]]

Output: 4

Explanation: One possible path is [0,1,4,2,3]

Constraints:

• n == graph.length
• 1 <= n <= 12
• 0 <= graph[i].length < n
• graph[i] does not contain i.
• If graph[a] contains b, then graph[b] contains a.
• The input graph is always connected.

Solution

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

class Solution {
    fun shortestPathLength(graph: Array<IntArray>): Int {
        val target = (1 shl graph.size) - 1
        val q: Queue<IntArray> = LinkedList()
        for (i in graph.indices) {
            q.offer(intArrayOf(i, 1 shl i))
        }
        var steps = 0
        val visited = Array(graph.size) {
            BooleanArray(
                target + 1
            )
        }
        while (q.isNotEmpty()) {
            val size = q.size
            for (i in 0 until size) {
                val curr = q.poll()
                val currNode = Objects.requireNonNull(curr)[0]
                val currState = curr[1]
                if (currState == target) {
                    return steps
                }
                for (n in graph[currNode]) {
                    val newState = currState or (1 shl n)
                    if (visited[n][newState]) {
                        continue
                    }
                    visited[n][newState] = true
                    q.offer(intArrayOf(n, newState))
                }
            }
            ++steps
        }
        return -1
    }
}

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