LeetCode #2360 — HARD

Longest Cycle in a Graph

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Topological Sort

Example 1

[3,3,4,2,3]

Example 2

[2,-1,3,1]

Core Insight

What unlocks the optimal approach

How many cycles can each node at most be part of?
Each node can be part of at most one cycle. Start from each node and find the cycle that it is part of if there is any. Save the already visited nodes to not repeat visiting the same cycle multiple times.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2360: Longest Cycle in a Graph
class Solution {
    public int longestCycle(int[] edges) {
        int n = edges.length;
        boolean[] vis = new boolean[n];
        int ans = -1;
        for (int i = 0; i < n; ++i) {
            if (vis[i]) {
                continue;
            }
            int j = i;
            List<Integer> cycle = new ArrayList<>();
            for (; j != -1 && !vis[j]; j = edges[j]) {
                vis[j] = true;
                cycle.add(j);
            }
            if (j == -1) {
                continue;
            }
            for (int k = 0; k < cycle.size(); ++k) {
                if (cycle.get(k) == j) {
                    ans = Math.max(ans, cycle.size() - k);
                    break;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2360: Longest Cycle in a Graph
func longestCycle(edges []int) int {
	vis := make([]bool, len(edges))
	ans := -1
	for i := range edges {
		if vis[i] {
			continue
		}
		j := i
		cycle := []int{}
		for ; j != -1 && !vis[j]; j = edges[j] {
			vis[j] = true
			cycle = append(cycle, j)
		}
		if j == -1 {
			continue
		}
		for k := range cycle {
			if cycle[k] == j {
				ans = max(ans, len(cycle)-k)
				break
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #2360: Longest Cycle in a Graph
class Solution:
    def longestCycle(self, edges: List[int]) -> int:
        n = len(edges)
        vis = [False] * n
        ans = -1
        for i in range(n):
            if vis[i]:
                continue
            j = i
            cycle = []
            while j != -1 and not vis[j]:
                vis[j] = True
                cycle.append(j)
                j = edges[j]
            if j == -1:
                continue
            m = len(cycle)
            k = next((k for k in range(m) if cycle[k] == j), inf)
            ans = max(ans, m - k)
        return ans

// Accepted solution for LeetCode #2360: Longest Cycle in a Graph
impl Solution {
    pub fn longest_cycle(edges: Vec<i32>) -> i32 {
        let n = edges.len();
        let mut vis = vec![false; n];
        let mut ans = -1;

        for i in 0..n {
            if vis[i] {
                continue;
            }
            let mut j = i as i32;
            let mut cycle = Vec::new();

            while j != -1 && !vis[j as usize] {
                vis[j as usize] = true;
                cycle.push(j);
                j = edges[j as usize];
            }

            if j == -1 {
                continue;
            }

            for k in 0..cycle.len() {
                if cycle[k] == j {
                    ans = ans.max((cycle.len() - k) as i32);
                    break;
                }
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #2360: Longest Cycle in a Graph
function longestCycle(edges: number[]): number {
    const n = edges.length;
    const vis: boolean[] = Array(n).fill(false);
    let ans = -1;
    for (let i = 0; i < n; ++i) {
        if (vis[i]) {
            continue;
        }
        let j = i;
        const cycle: number[] = [];
        for (; j !== -1 && !vis[j]; j = edges[j]) {
            vis[j] = true;
            cycle.push(j);
        }
        if (j === -1) {
            continue;
        }
        for (let k = 0; k < cycle.length; ++k) {
            if (cycle[k] === j) {
                ans = Math.max(ans, cycle.length - k);
                break;
            }
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

REPEATED DFS

O(V × E) time

O(V) space

Repeatedly find a vertex with no incoming edges, remove it and its outgoing edges, and repeat. Finding the zero-in-degree vertex scans all V vertices, and we do this V times. Removing edges touches E edges total. Without an in-degree array, this gives O(V × E).

TOPOLOGICAL SORT

O(V + E) time

O(V + E) space

Build an adjacency list (O(V + E)), then either do Kahn's BFS (process each vertex once + each edge once) or DFS (visit each vertex once + each edge once). Both are O(V + E). Space includes the adjacency list (O(V + E)) plus the in-degree array or visited set (O(V)).

Shortcut: Process each vertex once + each edge once → O(V + E). Same as BFS/DFS on a graph.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.