LeetCode in Kotlin

2360. Longest Cycle in a Graph

Hard

You are given a directed graph of n nodes numbered from 0 to n - 1, where each node has at most one outgoing edge.

The graph is represented with a given 0-indexed array edges of size n, indicating that there is a directed edge from node i to node edges[i]. If there is no outgoing edge from node i, then edges[i] == -1.

Return the length of the longest cycle in the graph. If no cycle exists, return -1.

A cycle is a path that starts and ends at the same node.

Example 1:

Input: edges = [3,3,4,2,3]

Output: 3

Explanation: The longest cycle in the graph is the cycle: 2 -> 4 -> 3 -> 2.

The length of this cycle is 3, so 3 is returned.

Example 2:

Input: edges = [2,-1,3,1]

Output: -1

Explanation: There are no cycles in this graph.

Constraints:

• n == edges.length
• 2 <= n <= 105
• -1 <= edges[i] < n
• edges[i] != i

Solution

class Solution {
    fun longestCycle(edges: IntArray): Int {
        val n = edges.size
        val vis = BooleanArray(n)
        val dfsvis = BooleanArray(n)
        val path = IntArray(n)
        var maxLength = -1
        for (i in 0 until n) {
            if (!vis[i]) {
                path[i] = 1
                maxLength = Math.max(maxLength, dfs(i, 1, path, vis, dfsvis, edges))
            }
        }
        return maxLength
    }

    private fun dfs(
        node: Int,
        pathLength: Int,
        path: IntArray,
        vis: BooleanArray,
        dfsvis: BooleanArray,
        edges: IntArray
    ): Int {
        vis[node] = true
        dfsvis[node] = true
        var length = -1
        if (edges[node] != -1 && !vis[edges[node]]) {
            path[edges[node]] = pathLength + 1
            length = dfs(edges[node], pathLength + 1, path, vis, dfsvis, edges)
        } else if (edges[node] != -1 && dfsvis[edges[node]]) {
            length = pathLength - path[edges[node]] + 1
        }
        dfsvis[node] = false
        return length
    }
}

This site is open source. Improve this page.