<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]]:
        # Step 1: Prepare the adjacency list for the graph
        graph = [[] for _ in range(n)]
        for u, v in connections:
            graph[u].append(v)
            graph[v].append(u)
        
        # Step 2: Initialize variables
        self.ids = [-1] * n  # id of each node
        self.low = [0] * n   # lowest id reachable from node
        self.id = 0          # initial node id
        self.bridges = []    # store critical connections
        self.visited = [False] * n
        
        # Step 3: Perform DFS to find bridges
        def dfs(at, parent):
            self.visited[at] = True
            self.ids[at] = self.low[at] = self.id
            self.id += 1
            
            # Visit all neighbors
            for to in graph[at]:
                if to == parent:
                    continue  # Skip the edge back to the parent
                
                if not self.visited[to]:
                    dfs(to, at)
                    # After dfs, check if the subtree rooted at 'to' has a connection back to 'at' or higher
                    self.low[at] = min(self.low[at], self.low[to])
                    
                    # A bridge is found if there is no back edge from 'to' to any ancestors of 'at'
                    if self.low[to] > self.ids[at]:
                        self.bridges.append([at, to])
                else:
                    # Update the low-link value on the callback
                    self.low[at] = min(self.low[at], self.ids[to])
        
        # Step 4: Start DFS from each node that hasn't been visited
        for i in range(n):
            if not self.visited[i]:
                dfs(i, -1)
        
        # Step 5: Return the bridges (critical connections)
        return self.bridges

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