What is the time complexity of Shortest Cycle in a Graph?

The optimal solution has a time complexity of O(n*(n+m)).

What is the space complexity of Shortest Cycle in a Graph?

The space complexity is O(n+m).

What data structures are used to solve Shortest Cycle in a Graph?

This problem uses Breadth-First Search, Graph. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Shortest Cycle in a Graph?

Insight 1: BFS is a suitable algorithm to determine the distance between nodes. We can use BFS from each node to explore the graph and find potential cycles. Insight 2: The key to identifying a cycle during BFS is to detect an adjacent node that has already been visited (i.e., has a distance value other than -1) and is not the parent node of the current node in the BFS traversal. This indicates a back edge, closing a cycle.

Which companies ask LeetCode 2608. Shortest Cycle in a Graph in interviews?

This problem is frequently asked at Zomato.

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Shortest Cycle in a Graph - LeetCode 2608 Solution

Shortest Cycle in a Graph - Complete Solution Guide

Shortest Cycle in a Graph is LeetCode problem 2608, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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

Detailed Explanation

The problem asks us to find the length of the shortest cycle in a given undirected graph. A cycle is a path that starts and ends at the same node, and each edge in the path is used only once. The graph is represented by a list of edges, where each edge connects two vertices. The goal is to return the minimum length of any cycle present in the graph. If the graph doesn't contain any cycles, we should return -1. The graph has 'n' vertices labeled from 0 to n-1. There are no self-loops (edge from a vertex to itself) and no repeated edges.

Solution Approach

The solution uses Breadth-First Search (BFS) to find the shortest cycle in the graph. For each node in the graph, BFS is performed to explore its neighbors. During the BFS traversal, we keep track of the distance from the starting node and the parent of each visited node. If we encounter a visited neighbor that is not the parent of the current node, we've found a cycle. The length of this cycle is the sum of the distances from the starting node to the two nodes plus 1 (for the edge connecting the two nodes). The algorithm iterates through all nodes, performing BFS from each and updating the minimum cycle length found so far.

Step-by-Step Algorithm

Step 1: Create an adjacency list representation of the graph from the given edges. This allows efficient access to a node's neighbors.
Step 2: Initialize the minimum cycle length to infinity (or a very large number) to ensure that any cycle found will be shorter.
Step 3: Iterate through each node 'i' from 0 to n-1. This node will be the starting point for a BFS traversal.
Step 4: For each node 'i', perform BFS: Initialize distance and parent arrays. Distance array stores the distance from node 'i' to other nodes. Parent array stores the parent of each node in the BFS tree.
Step 5: Enqueue the starting node 'i' into the BFS queue and set its distance to 0.
Step 6: While the queue is not empty, dequeue a node 'u'.
Step 7: Iterate through each neighbor 'v' of node 'u'.
Step 8: If 'v' has not been visited (distance[v] == -1), update its distance, set its parent to 'u', and enqueue 'v'.
Step 9: If 'v' has been visited and 'v' is not the parent of 'u', a cycle is detected. The length of the cycle is distance[u] + distance[v] + 1. Update the minimum cycle length if the detected cycle is shorter.
Step 10: After BFS from node 'i' completes, proceed to the next node and repeat steps 4-9.
Step 11: After performing BFS from all nodes, if the minimum cycle length is still infinity, there are no cycles in the graph. Return -1. Otherwise, return the minimum cycle length found.

Key Insights

Insight 1: BFS is a suitable algorithm to determine the distance between nodes. We can use BFS from each node to explore the graph and find potential cycles.
Insight 2: The key to identifying a cycle during BFS is to detect an adjacent node that has already been visited (i.e., has a distance value other than -1) and is not the parent node of the current node in the BFS traversal. This indicates a back edge, closing a cycle.
Insight 3: Since we need to find the *shortest* cycle, we need to perform BFS from every node and track the minimum cycle length found so far. This ensures that we explore all possible cycles in the graph.

Complexity Analysis

Time Complexity: O(n*(n+m))

Space Complexity: O(n+m)

Topics

This problem involves: Breadth-First Search, Graph.

Companies

Asked at: Zomato.

Shortest Cycle in a Graph | LeetCode 2608

Hard

Breadth-First Search Graph

Time: O(n*(n+m))Space: O(n+m)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;

class Solution {
    public int findShortestCycle(int n, int[][] edges) {
        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adj.get(u).add(v);
            adj.get(v).add(u);
        }

        int min_len = Integer.MAX_VALUE;

        for (int i = 0; i < n; i++) {
            int[] dist = new int[n];
            Arrays.fill(dist, -1);
            int[] parent = new int[n];
            Arrays.fill(parent, -1);
            Queue<Integer> q = new LinkedList<>();
            q.offer(i);
            dist[i] = 0;

            while (!q.isEmpty()) {
                int u = q.poll();
                for (int v : adj.get(u)) {
                    if (dist[v] == -1) {
                        dist[v] = dist[u] + 1;
                        parent[v] = u;
                        q.offer(v);
                    } else if (v != parent[u]) {
                        min_len = Math.min(min_len, dist[u] + dist[v] + 1);
                    }
                }
            }
        }

        if (min_len == Integer.MAX_VALUE) {
            return -1;
        } else {
            return min_len;
        }
    }
}

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Problem Info

Description

Problem 2608: Shortest Cycle in a Graph - Detailed explanation and solution approaches.

Topics

Breadth-First Search Graph

Companies

Zomato

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2608

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Create an adjacency list representation of the graph from the given edges. This allows efficient access to a node's neighbors.
2
Step 2: Initialize the minimum cycle length to infinity (or a very large number) to ensure that any cycle found will be shorter.
3
Step 3: Iterate through each node 'i' from 0 to n-1. This node will be the starting point for a BFS traversal.
4
Step 4: For each node 'i', perform BFS: Initialize distance and parent arrays. Distance array stores the distance from node 'i' to other nodes. Parent array stores the parent of each node in the BFS tree.
5
Step 5: Enqueue the starting node 'i' into the BFS queue and set its distance to 0.
6
Step 6: While the queue is not empty, dequeue a node 'u'.
7
Step 7: Iterate through each neighbor 'v' of node 'u'.
8
Step 8: If 'v' has not been visited (distance[v] == -1), update its distance, set its parent to 'u', and enqueue 'v'.
9
Step 9: If 'v' has been visited and 'v' is not the parent of 'u', a cycle is detected. The length of the cycle is distance[u] + distance[v] + 1. Update the minimum cycle length if the detected cycle is shorter.
10
Step 10: After BFS from node 'i' completes, proceed to the next node and repeat steps 4-9.
11
Step 11: After performing BFS from all nodes, if the minimum cycle length is still infinity, there are no cycles in the graph. Return -1. Otherwise, return the minimum cycle length found.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*(n+m))

💾

Space Complexity

O(n+m)

📝

Why?

The time complexity is derived from performing a BFS from each of the 'n' nodes in the graph. Each BFS can potentially visit all 'n' nodes and 'm' edges. The space complexity is determined by the storage required for the adjacency list (which stores the graph structure), the distance array (used to track distances during BFS), the parent array (used to track the path), and the BFS queue itself.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Depth First Search (DFS)

DFS can also be used to detect cycles. The idea is to keep track of the nodes in the current path. If we encounter a visited node that is already in the current path, we have found a cycle. We would need to track distances and potentially the full path taken so far to get the cycle length. The shortest cycle can be obtained by performing DFS starting from each node.

Time: O(n * (n + m))Space: O(n + m)

Trade-offs: While DFS also has a time complexity of O(n*(n+m)), BFS tends to find shorter cycles more quickly. BFS also requires more memory due to the queue, DFS requires less.

Floyd-Warshall Algorithm

The Floyd-Warshall algorithm can find the shortest paths between all pairs of nodes in a graph. If the graph has any cycles, the shortest path from a node to itself will be a cycle. After running Floyd-Warshall, the minimum non-zero value on the main diagonal of the distance matrix will be the length of the shortest cycle.

Time: O(n^3)Space: O(n^2)

Trade-offs: Floyd-Warshall is simpler to implement for this specific problem, but has a higher time complexity than BFS, and requires much more space. For sparse graphs, the BFS solution will likely be faster. Floyd Warshall is suitable for dense graphs.

✨

Key Insights

1Insight 1: BFS is a suitable algorithm to determine the distance between nodes. We can use BFS from each node to explore the graph and find potential cycles.
2Insight 2: The key to identifying a cycle during BFS is to detect an adjacent node that has already been visited (i.e., has a distance value other than -1) and is not the parent node of the current node in the BFS traversal. This indicates a back edge, closing a cycle.
3Insight 3: Since we need to find the *shortest* cycle, we need to perform BFS from every node and track the minimum cycle length found so far. This ensures that we explore all possible cycles in the graph.

📚

Key Terms Explained

For Students

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: Graph with no edges. The code correctly returns -1 because the `min_len` will remain at infinity/Integer.MAX_VALUE.
Edge case 2: Disconnected graph. The code correctly finds the shortest cycle within each connected component because it performs BFS starting from every node. If a component is acyclic, the BFS starting from that component's nodes won't find any cycle.
Edge case 3: Graph with only one edge. The code correctly returns -1, as a single edge cannot form a cycle.
Edge case 4: Graph is a tree (no cycles). The BFS from each node will not find any valid cycle (the `v != parent[u]` condition will always fail when `dist[v] != -1`).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the parent node correctly. Failing to check if the adjacent node 'v' is the parent node leads to incorrectly identifying non-cycle paths as cycles.

✗Common mistake 2: Not considering all starting nodes. The algorithm must perform BFS starting from every node to ensure that the shortest cycle is found. Stopping after finding the first cycle might not yield the shortest one.

✗Common mistake 3: Incorrectly calculating cycle length. The length of the cycle is the sum of distances to each node in the cycle from the BFS starting node *plus one* for the final edge that completes the cycle.

✗Common mistake 4: Integer Overflow: When calculating cycle lengths with large distances, adding 1 to a large integer may cause an overflow, particularly in C/C++. Using a `long` type may prevent this problem.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Network Routing: Finding the shortest cycle in a network topology can help in designing efficient routing protocols and avoiding routing loops.

🏦Real-world scenario 2: Chemical Structure Analysis: Identifying cycles in molecular graphs is crucial for understanding chemical properties and reactions. The shortest cycle can represent the smallest ring structure in a molecule.

🎮Real-world scenario 3: Detecting Circular Dependencies in Software: Cycles in dependency graphs indicate circular dependencies between software modules, which can lead to build issues or runtime errors. Finding and breaking these cycles is crucial for maintainable software design.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 28, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

How do I represent a graph in code?

Use adjacency list (HashMap or array of lists) for sparse graphs - it's space-efficient O(V+E). Use adjacency matrix for dense graphs or when you need O(1) edge lookup. For weighted graphs, store (neighbor, weight) pairs in adjacency lists.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Shortest Cycle in a Graph"?

For this hard-level Breadth-First Search problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether graph techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?