LeetCode #2608 — HARD

Shortest Cycle in a Graph

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

The Problem

Problem Statement

There is a bi-directional graph with n vertices, where each vertex is labeled from 0 to n - 1. The edges in the graph are represented by a given 2D integer array edges, where edges[i] = [u_i, v_i] denotes an edge between vertex u_i and vertex v_i. Every vertex pair is connected by at most one edge, and no vertex has an edge to itself.

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

A cycle is a path that starts and ends at the same node, and each edge in the path is used only once.

Example 1:

Input: n = 7, edges = [[0,1],[1,2],[2,0],[3,4],[4,5],[5,6],[6,3]]
Output: 3
Explanation: The cycle with the smallest length is : 0 -> 1 -> 2 -> 0

Example 2:

Input: n = 4, edges = [[0,1],[0,2]]
Output: -1
Explanation: There are no cycles in this graph.

Constraints:

2 <= n <= 1000
1 <= edges.length <= 1000
edges[i].length == 2
0 <= u_i, v_i < n
u_i != v_i
There are no repeated edges.

Step 01

Brute Force Baseline

Problem summary: There is a bi-directional graph with n vertices, where each vertex is labeled from 0 to n - 1. The edges in the graph are represented by a given 2D integer array edges, where edges[i] = [ui, vi] denotes an edge between vertex ui and vertex vi. Every vertex pair is connected by at most one edge, and no vertex has an edge to itself. Return the length of the shortest cycle in the graph. If no cycle exists, return -1. A cycle is a path that starts and ends at the same node, and each edge in the path is used only once.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

7
[[0,1],[1,2],[2,0],[3,4],[4,5],[5,6],[6,3]]

Example 2

4
[[0,1],[0,2]]

Core Insight

What unlocks the optimal approach

How can BFS be used?
For each vertex u, calculate the length of the shortest cycle that contains vertex u using BFS

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 #2608: Shortest Cycle in a Graph
class Solution {
    private List<Integer>[] g;
    private final int inf = 1 << 30;

    public int findShortestCycle(int n, int[][] edges) {
        g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : edges) {
            int u = e[0], v = e[1];
            g[u].add(v);
            g[v].add(u);
        }
        int ans = inf;
        for (var e : edges) {
            int u = e[0], v = e[1];
            ans = Math.min(ans, bfs(u, v));
        }
        return ans < inf ? ans : -1;
    }

    private int bfs(int u, int v) {
        int[] dist = new int[g.length];
        Arrays.fill(dist, inf);
        dist[u] = 0;
        Deque<Integer> q = new ArrayDeque<>();
        q.offer(u);
        while (!q.isEmpty()) {
            int i = q.poll();
            for (int j : g[i]) {
                if ((i == u && j == v) || (i == v && j == u) || dist[j] != inf) {
                    continue;
                }
                dist[j] = dist[i] + 1;
                q.offer(j);
            }
        }
        return dist[v] + 1;
    }
}

// Accepted solution for LeetCode #2608: Shortest Cycle in a Graph
func findShortestCycle(n int, edges [][]int) int {
	g := make([][]int, n)
	for _, e := range edges {
		u, v := e[0], e[1]
		g[u] = append(g[u], v)
		g[v] = append(g[v], u)
	}
	const inf = 1 << 30
	bfs := func(u, v int) int {
		dist := make([]int, n)
		for i := range dist {
			dist[i] = inf
		}
		dist[u] = 0
		q := []int{u}
		for len(q) > 0 {
			i := q[0]
			q = q[1:]
			for _, j := range g[i] {
				if (i == u && j == v) || (i == v && j == u) || dist[j] != inf {
					continue
				}
				dist[j] = dist[i] + 1
				q = append(q, j)
			}
		}
		return dist[v] + 1
	}
	ans := inf
	for _, e := range edges {
		u, v := e[0], e[1]
		ans = min(ans, bfs(u, v))
	}
	if ans < inf {
		return ans
	}
	return -1
}

# Accepted solution for LeetCode #2608: Shortest Cycle in a Graph
class Solution:
    def findShortestCycle(self, n: int, edges: List[List[int]]) -> int:
        def bfs(u: int, v: int) -> int:
            dist = [inf] * n
            dist[u] = 0
            q = deque([u])
            while q:
                i = q.popleft()
                for j in g[i]:
                    if (i, j) != (u, v) and (j, i) != (u, v) and dist[j] == inf:
                        dist[j] = dist[i] + 1
                        q.append(j)
            return dist[v] + 1

        g = defaultdict(set)
        for u, v in edges:
            g[u].add(v)
            g[v].add(u)
        ans = min(bfs(u, v) for u, v in edges)
        return ans if ans < inf else -1

// Accepted solution for LeetCode #2608: Shortest Cycle in a Graph
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2608: Shortest Cycle in a Graph
// class Solution {
//     private List<Integer>[] g;
//     private final int inf = 1 << 30;
// 
//     public int findShortestCycle(int n, int[][] edges) {
//         g = new List[n];
//         Arrays.setAll(g, k -> new ArrayList<>());
//         for (var e : edges) {
//             int u = e[0], v = e[1];
//             g[u].add(v);
//             g[v].add(u);
//         }
//         int ans = inf;
//         for (var e : edges) {
//             int u = e[0], v = e[1];
//             ans = Math.min(ans, bfs(u, v));
//         }
//         return ans < inf ? ans : -1;
//     }
// 
//     private int bfs(int u, int v) {
//         int[] dist = new int[g.length];
//         Arrays.fill(dist, inf);
//         dist[u] = 0;
//         Deque<Integer> q = new ArrayDeque<>();
//         q.offer(u);
//         while (!q.isEmpty()) {
//             int i = q.poll();
//             for (int j : g[i]) {
//                 if ((i == u && j == v) || (i == v && j == u) || dist[j] != inf) {
//                     continue;
//                 }
//                 dist[j] = dist[i] + 1;
//                 q.offer(j);
//             }
//         }
//         return dist[v] + 1;
//     }
// }

// Accepted solution for LeetCode #2608: Shortest Cycle in a Graph
function findShortestCycle(n: number, edges: number[][]): number {
    const g: number[][] = new Array(n).fill(0).map(() => []);
    for (const [u, v] of edges) {
        g[u].push(v);
        g[v].push(u);
    }
    const inf = 1 << 30;
    let ans = inf;
    const bfs = (u: number, v: number) => {
        const dist: number[] = new Array(n).fill(inf);
        dist[u] = 0;
        const q: number[] = [u];
        while (q.length) {
            const i = q.shift()!;
            for (const j of g[i]) {
                if ((i == u && j == v) || (i == v && j == u) || dist[j] != inf) {
                    continue;
                }
                dist[j] = dist[i] + 1;
                q.push(j);
            }
        }
        return 1 + dist[v];
    };
    for (const [u, v] of edges) {
        ans = Math.min(ans, bfs(u, v));
    }
    return ans < inf ? ans : -1;
}

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(m^2)

Space

O(m + n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.