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Leetcode #2608: Shortest Cycle in a Graph

In this guide, we solve Leetcode #2608 Shortest Cycle in a Graph in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

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] = [ui, vi] denotes an edge between vertex ui and vertex vi.

Quick Facts

  • Difficulty: Hard
  • Premium: No
  • Tags: Breadth-First Search, Graph

Intuition

The data forms a graph, so we should explore nodes and edges systematically.

A traversal ensures we visit each node once while maintaining the needed state.

Approach

Build an adjacency list and traverse with BFS or DFS.

Aggregate results as you visit nodes.

Steps:

  • Build the graph.
  • Traverse with BFS/DFS.
  • Accumulate the required output.

Example

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

Python Solution

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

Complexity

The time complexity is O(m2)O(m^2)O(m2) and the space complexity is O(m+n)O(m + n)O(m+n), where mmm and nnn are the length of the array edgesedgesedges and the number of vertices. The space complexity is O(m+n)O(m + n)O(m+n), where mmm and nnn are the length of the array edgesedgesedges and the number of vertices.

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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