How do you solve Critical Connections in a Network on LeetCode?

Critical Connections in a Network (LeetCode #1192) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to track discovery and low values is crucial for identifying critical connections.

What is the brute force solution for Critical Connections in a Network?

The brute force approach for Critical Connections in a Network is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every connection and simulates its removal to see if the network becomes disconnected. While simple, this method is inefficient for larger networks. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Critical Connections in a Network?

Critical Connections in a Network uses the following concepts: Depth-First Search, Graph Theory, Biconnected Component. The recommended patterns to study are: Depth-First Search, Graph Theory, Biconnected Components.

What are the interview tips for Critical Connections in a Network?

Explain your thought process clearly while coding; it shows your understanding. Be prepared to discuss the time and space complexities of your approach. Practice explaining Tarjan's algorithm as it's a common topic in graph-related problems.

What are common mistakes when solving Critical Connections in a Network?

Failing to reset the parent array or discovery/low values between different DFS calls. Not considering back edges correctly when updating low values.

#1192

Critical Connections in a Network

Hard

Depth-First SearchGraph TheoryBiconnected ComponentDepth-First SearchGraph TheoryBiconnected Components

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using Tarjan's algorithm, we can efficiently find critical connections in O(n) time by leveraging depth-first search and tracking discovery and low-link values.

⚙️

Algorithm

4 steps

1Step 1: Build the graph using an adjacency list.
2Step 2: Initialize discovery and low arrays, and a parent array to track the DFS tree.
3Step 3: Perform DFS, updating discovery and low values to identify critical connections.
4Step 4: If the low value of a neighbor is greater than the discovery value of the current node, it's a critical connection.

solution.py27 lines

1def criticalConnections(n, connections):
2    from collections import defaultdict
3    graph = defaultdict(list)
4    for a, b in connections:
5        graph[a].append(b)
6        graph[b].append(a)
7    discovery = [-1] * n
8    low = [-1] * n
9    parent = [-1] * n
10    critical = []
11    time = [0]
12    def dfs(node):
13        discovery[node] = low[node] = time[0]
14        time[0] += 1
15        for neighbor in graph[node]:
16            if discovery[neighbor] == -1:
17                parent[neighbor] = node
18                dfs(neighbor)
19                low[node] = min(low[node], low[neighbor])
20                if low[neighbor] > discovery[node]:
21                    critical.append([node, neighbor])
22            elif neighbor != parent[node]:
23                low[node] = min(low[node], discovery[neighbor])
24    for i in range(n):
25        if discovery[i] == -1:
26            dfs(i)
27    return critical

ℹ

Complexity note: The time complexity is O(n) because we visit each node and edge once during the DFS traversal. The space complexity is O(n) due to the storage of the graph and auxiliary arrays.

1Understanding how to track discovery and low values is crucial for identifying critical connections.
2Recognizing the importance of graph traversal techniques like DFS in solving connectivity problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.