What is the time complexity of Critical Connections in a Network?

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

What is the space complexity of Critical Connections in a Network?

The space complexity is O(n+e).

What data structures are used to solve Critical Connections in a Network?

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

What are the key insights for solving Critical Connections in a Network?

Insight 1: The core idea is to use Depth First Search (DFS) to explore the graph and identify bridges using Tarjan's algorithm. This involves tracking discovery times and low-link values. Insight 2: Discovery time records when a node is first visited during the DFS traversal. Low-link value represents the earliest visited node reachable from the current node through the DFS subtree, including back edges.

Which companies ask LeetCode 1192. Critical Connections in a Network in interviews?

This problem is frequently asked at Akuna Capital.

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Critical Connections in a Network - LeetCode 1192 Solution

Critical Connections in a Network - Complete Solution Guide

Critical Connections in a Network is LeetCode problem 1192, 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 are n servers numbered from 0 to n - 1 connected by undirected server-to-server connections forming a network where connections[i] = [a i , b i ] represents a connection between servers a i and b i . Any server can reach other servers directly or indirectly through the network. A critical connection is a connection that, if removed, will make some servers unable to reach some other server. Return all critical connections in the network in any order. Example 1: Input: n = 4, connections = [

Detailed Explanation

The problem asks us to find all critical connections (also known as bridges) in a given undirected graph. A critical connection is an edge that, if removed, would increase the number of connected components in the graph. In simpler terms, it's an edge whose removal would disconnect part of the graph from the rest.

Solution Approach

The provided solutions implement Tarjan's bridge-finding algorithm using Depth First Search (DFS). The algorithm works by assigning a discovery time and a low-link value to each node. The discovery time is the time when the node is first visited during the DFS traversal. The low-link value is the smallest discovery time of any node reachable from the current node through the DFS subtree (including back edges). If the low-link value of a child node `v` is greater than the discovery time of its parent node `u`, then the edge `(u, v)` is a bridge.

Step-by-Step Algorithm

Step 1: Build the graph representation using an adjacency list. Each node stores a list of its neighbors.
Step 2: Initialize arrays to store discovery times and low-link values for each node. Initialize them to -1 to indicate that the nodes haven't been visited yet.
Step 3: Initialize a time variable to 0. This variable will keep track of the discovery time.
Step 4: Iterate through each node in the graph. If a node hasn't been visited yet (discovery time is -1), start a DFS traversal from that node.
Step 5: During the DFS traversal, for each node `u`, assign its discovery time and low-link value to the current time, and then increment the time.
Step 6: Iterate through each neighbor `v` of node `u`. If `v` is the parent of `u`, skip it to avoid going back up the DFS tree immediately. Also, prevents treating u->parent as a backedge.
Step 7: If `v` hasn't been visited yet (discovery time is -1), recursively call DFS on `v`. After the recursive call returns, update the low-link value of `u` to the minimum of its current low-link value and the low-link value of `v`. If the low-link value of `v` is greater than the discovery time of `u`, then the edge `(u, v)` is a bridge and add it to the result list.
Step 8: If `v` has already been visited, update the low-link value of `u` to the minimum of its current low-link value and the discovery time of `v`. This step considers back edges.
Step 9: After the DFS traversal is complete, return the list of bridges.

Key Insights

Insight 1: The core idea is to use Depth First Search (DFS) to explore the graph and identify bridges using Tarjan's algorithm. This involves tracking discovery times and low-link values.
Insight 2: Discovery time records when a node is first visited during the DFS traversal. Low-link value represents the earliest visited node reachable from the current node through the DFS subtree, including back edges.
Insight 3: An edge (u, v) is a bridge if and only if the low-link value of v is greater than the discovery time of u. This means there's no path back to u (or an ancestor of u) from v other than through the edge (u, v).
Insight 4: We need to keep track of the parent of each node during the DFS to avoid treating the edge to the parent as a back edge incorrectly.

Complexity Analysis

Time Complexity: O(n+e)

Space Complexity: O(n+e)

Topics

This problem involves: Depth-First Search, Graph, Biconnected Component.

Companies

Asked at: Akuna Capital.

Critical Connections in a Network | LeetCode 1192

Hard

Depth-First Search Graph Biconnected Component

Time: O(n+e)Space: O(n+e)

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<List<Integer>> criticalConnections(int n, List<List<Integer>> connections) {
        Map<Integer, List<Integer>> graph = new HashMap<>();
        for (int i = 0; i < n; i++) {
            graph.put(i, new ArrayList<>());
        }
        for (List<Integer> connection : connections) {
            int u = connection.get(0);
            int v = connection.get(1);
            graph.get(u).add(v);
            graph.get(v).add(u);
        }

        int[] discovery_times = new int[n];
        Arrays.fill(discovery_times, -1);
        int[] low_links = new int[n];
        Arrays.fill(low_links, -1);
        int time = 0;
        List<List<Integer>> result = new ArrayList<>();

        class DFS {
            void find_bridges_dfs(int u, int parent) {
                discovery_times[u] = low_links[u] = time;
                time += 1;

                for (int v : graph.get(u)) {
                    if (v == parent) {
                        continue;
                    }

                    if (discovery_times[v] == -1) {
                        find_bridges_dfs(v, u);
                        low_links[u] = Math.min(low_links[u], low_links[v]);
                        if (low_links[v] > discovery_times[u]) {
                            result.add(Arrays.asList(u, v));
                        }
                    } else {
                        low_links[u] = Math.min(low_links[u], discovery_times[v]);
                    }
                }
            }
        }

        DFS dfsInstance = new DFS();

        for (int i = 0; i < n; i++) {
            if (discovery_times[i] == -1) {
                dfsInstance.find_bridges_dfs(i, -1);
            }
        }

        return result;
    }
}

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Problem Info

Description

Problem 1192: Critical Connections in a Network - Detailed explanation and solution approaches.

Topics

Depth-First Search Graph Biconnected Component

Companies

Akuna Capital

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1192

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

9 steps

1
Step 1: Build the graph representation using an adjacency list. Each node stores a list of its neighbors.
2
Step 2: Initialize arrays to store discovery times and low-link values for each node. Initialize them to -1 to indicate that the nodes haven't been visited yet.
3
Step 3: Initialize a time variable to 0. This variable will keep track of the discovery time.
4
Step 4: Iterate through each node in the graph. If a node hasn't been visited yet (discovery time is -1), start a DFS traversal from that node.
5
Step 5: During the DFS traversal, for each node `u`, assign its discovery time and low-link value to the current time, and then increment the time.
6
Step 6: Iterate through each neighbor `v` of node `u`. If `v` is the parent of `u`, skip it to avoid going back up the DFS tree immediately. Also, prevents treating u->parent as a backedge.
7
Step 7: If `v` hasn't been visited yet (discovery time is -1), recursively call DFS on `v`. After the recursive call returns, update the low-link value of `u` to the minimum of its current low-link value and the low-link value of `v`. If the low-link value of `v` is greater than the discovery time of `u`, then the edge `(u, v)` is a bridge and add it to the result list.
8
Step 8: If `v` has already been visited, update the low-link value of `u` to the minimum of its current low-link value and the discovery time of `v`. This step considers back edges.
9
Step 9: After the DFS traversal is complete, return the list of bridges.

💡

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+e)

💾

Space Complexity

O(n+e)

📝

Why?

Building the graph takes O(e) time. DFS traversal visits each node and edge once, contributing O(n+e). The arrays to track discovery times and low links take O(n) space. The recursive nature of DFS may, in the worst case, lead to O(n) stack frames.

📖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

Path Finding Algorithms (e.g. Dijkstra, BFS)

One could, theoretically, perform a graph traversal for all node pairs and determine if alternative paths exist. If only one path exists between any node pair, the edge forming that path would be a critical connection.

Time: O(e * (n+e)) if doing BFS/DFS for each edge to verify connectivity after removal of that edge.Space: O(n+e)

Trade-offs: This approach is significantly less efficient than Tarjan's algorithm. Tarjan's algorithm identifies critical connections in a single DFS traversal, while the alternative approaches require multiple traversals for each possible edge.

✨

Key Insights

1Insight 1: The core idea is to use Depth First Search (DFS) to explore the graph and identify bridges using Tarjan's algorithm. This involves tracking discovery times and low-link values.
2Insight 2: Discovery time records when a node is first visited during the DFS traversal. Low-link value represents the earliest visited node reachable from the current node through the DFS subtree, including back edges.
3Insight 3: An edge (u, v) is a bridge if and only if the low-link value of v is greater than the discovery time of u. This means there's no path back to u (or an ancestor of u) from v other than through the edge (u, v).
4Insight 4: We need to keep track of the parent of each node during the DFS to avoid treating the edge to the parent as a back edge incorrectly.

📚

Key Terms Explained

For Students

Graph

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

Edge Cases

Edge case 1: Disconnected graph. The algorithm handles this correctly because it iterates through all nodes and starts a DFS traversal from any unvisited node, thus visiting all connected components.
Edge case 2: A graph with no edges. In this case, there will be no critical connections, and the algorithm correctly returns an empty list.
Edge case 3: A graph with only one edge. That single edge will be identified as a critical connection.
Edge case 4: A complete graph. Complete graphs do not have any critical connections (bridges) as any edge removal will not disconnect the graph. The algorithm will correctly return an empty list.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to skip the parent node in the DFS traversal. This can lead to incorrect low-link value calculations and misidentification of edges as bridges.

✗Common mistake 2: Incorrectly updating the low-link value. The low-link value should be updated with the minimum of its current value and the discovery time of the neighbor (for back edges) or the low-link value of the child (after recursive call).

✗Common mistake 3: Not handling disconnected graphs properly. The DFS traversal needs to be started from each unvisited node to ensure all connected components are visited.

✗Common mistake 4: Not understanding the difference between discovery time and low-link value. These are critical to the correctness of Tarjan's algorithm.

✅

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 infrastructure analysis. Identifying critical connections in a network can help ensure that the network remains connected even if some connections fail. This can be used to design more resilient networks.

🏦Real-world scenario 2: Supply chain management. Critical connections in a supply chain can be identified to ensure that the supply chain remains operational even if some links are disrupted. For example, identifying a single supplier who is the only source for a critical component.

🎮Real-world scenario 3: Power grid analysis. Critical connections in a power grid can be identified to ensure that the power grid remains stable even if some lines are taken out of service.

💡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: September 21, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

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

@CodeNinjaJul 6

❓

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 "Critical Connections in a Network"?

For this hard-level Depth-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?