Introduction

In this blog post, we will explore how to solve the LeetCode problem "1192" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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] = [ai, bi] represents a connection between servers ai and bi. 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 = [[0,1],[1,2],[2,0],[1,3]] Output: [[1,3]] Explanation: [[3,1]] is also accepted. Example 2: Input: n = 2, connections = [[0,1]] Output: [[0,1]] Constraints: 2 <= n <= 105 n - 1 <= connections.length <= 105 0 <= ai, bi <= n - 1 ai != bi There are no repeated connections.

Explanation

Here's a breakdown of the solution:

• Tarjan's Algorithm: The core idea is to use Tarjan's algorithm to find bridges (critical connections) in the graph. Tarjan's algorithm is an efficient way to find strongly connected components and articulation points/bridges in a graph.

• Depth-First Search (DFS): A DFS is performed, keeping track of the discovery time (disc) and lowest reachable ancestor (low) for each node. Bridges are identified when disc[u] < low[v] for an edge (u, v). This means v cannot reach u or any ancestor of u without going through the edge (u, v), hence it's a critical connection.

• Efficient Implementation: The graph is represented using an adjacency list for fast neighbor lookups. disc and low arrays are used to efficiently track discovery times and lowest reachable ancestors during the DFS.

• Time & Space Complexity: O(V+E) for both time and space, where V is the number of vertices (servers) and E is the number of edges (connections).

Code

    def criticalConnections(n: int, connections: list[list[int]]) -> list[list[int]]:
    """
    Finds all critical connections (bridges) in a network of servers.

    Args:
        n: The number of servers.
        connections: A list of connections between servers.

    Returns:
        A list of critical connections.
    """

    graph = [[] for _ in range(n)]
    for u, v in connections:
        graph[u].append(v)
        graph[v].append(u)

    disc = [-1] * n  # Discovery time of each node
    low = [-1] * n  # Lowest reachable ancestor of each node
    parent = [-1] * n  # Parent of each node in DFS tree
    critical_connections = []
    time = 0

    def dfs(u):
        nonlocal time
        disc[u] = time
        low[u] = time
        time += 1

        for v in graph[u]:
            if disc[v] == -1:  # If v is not visited
                parent[v] = u
                dfs(v)

                low[u] = min(low[u], low[v])

                if disc[u] < low[v]:
                    critical_connections.append([u, v])
            elif v != parent[u]:  # Ignore the parent edge
                low[u] = min(low[u], disc[v])

    for i in range(n):
        if disc[i] == -1:
            dfs(i)

    return critical_connections