<p>There are <code>n</code> servers numbered from <code>0</code> to <code>n - 1</code> connected by undirected server-to-server <code>connections</code> forming a network where <code>connections[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> represents a connection between servers <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code>. Any server can reach other servers directly or indirectly through the network.</p>

<p>A <em>critical connection</em> is a connection that, if removed, will make some servers unable to reach some other server.</p>

<p>Return all critical connections in the network in any order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/09/03/1537_ex1_2.png" style="width: 198px; height: 248px;" />
<pre>
<strong>Input:</strong> n = 4, connections = [[0,1],[1,2],[2,0],[1,3]]
<strong>Output:</strong> [[1,3]]
<strong>Explanation:</strong> [[3,1]] is also accepted.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 2, connections = [[0,1]]
<strong>Output:</strong> [[0,1]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>n - 1 &lt;= connections.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li>There are no repeated connections.</li>
</ul>

class Solution:
    def criticalConnections(self, n: int, connections: list[list[int]]) -> list[list[int]]:
        from collections import defaultdict
        
        # Create adjacency list
        graph = defaultdict(list)
        for u, v in connections:
            graph[u].append(v)
            graph[v].append(u)
        
        # Arrays to store discovery times and the lowest reachable vertex
        disc = [-1] * n
        low = [-1] * n
        visited = [False] * n
        bridges = []
        time = [0]  # Mutable container to keep track of time
        
        def dfs(u: int, parent: int):
            visited[u] = True
            disc[u] = low[u] = time[0]
            time[0] += 1
            
            for v in graph[u]:
                if not visited[v]:  # v is not visited
                    dfs(v, u)
                    low[u] = min(low[u], low[v])
                    if low[v] > disc[u]:
                        bridges.append([u, v])
                elif v != parent:  # Back edge
                    low[u] = min(low[u], disc[v])
        
        # Call the DFS function for each component
        for i in range(n):
            if not visited[i]:
                dfs(i, -1)
        
        return bridges

74666