<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]]:
        # Create a graph using an adjacency list
        graph = defaultdict(list)
        for u, v in connections:
            graph[u].append(v)
            graph[v].append(u)
        
        # Helper variables
        discovery = [-1] * n  # Discovery times of visited vertices
        low = [-1] * n  # Lowest discovery time reachable from a vertex
        result = []  # List to store the critical connections (bridges)
        time = 0  # Global timer
        
        # DFS function
        def dfs(node: int, parent: int):
            nonlocal time
            discovery[node] = low[node] = time
            time += 1
            
            # Traverse all neighbors of the current node
            for neighbor in graph[node]:
                if neighbor == parent:
                    # If the neighbor is the parent, skip it
                    continue
                if discovery[neighbor] == -1:
                    # If neighbor has not been visited, recurse on it
                    dfs(neighbor, node)
                    # Check if the subtree rooted at neighbor can reach an ancestor of node
                    low[node] = min(low[node], low[neighbor])
                    
                    # If the lowest point reachable from neighbor is above discovery[node],
                    # this edge is a critical connection (bridge)
                    if low[neighbor] > discovery[node]:
                        result.append([node, neighbor])
                else:
                    # If neighbor has been visited and is not the parent, update low[node]
                    low[node] = min(low[node], discovery[neighbor])
        
        # Start DFS from node 0 (assuming the graph is connected)
        for i in range(n):
            if discovery[i] == -1:
                dfs(i, -1)
        
        return result
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