<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>

from collections import defaultdict
class Solution:
    def criticalConnections(self, n: int, connections: List[List[int]]) -> List[List[int]]:
        """
        - graph is fully connected
        - our goal is to remove an edge/connection such that the graph becomes disconnected
        - specifically, remove critical connections
        """
        adj = defaultdict(list)
        self.time = 0
        lowlink = [-1] * n
        ids = [-1] * n
        visited = set([])
        bridges = []

        for u, v in connections:
            adj[u].append(v)
            adj[v].append(u)

        def dfs(node, parent):
            visited.add(node)
            ids[node] = lowlink[node] = self.time
            self.time += 1
            for neighbor in adj[node]:
                if neighbor == parent:
                    continue

                if neighbor not in visited:
                    dfs(neighbor, node)
                    lowlink[node] = min(lowlink[neighbor], lowlink[node])
                
                    if lowlink[neighbor] > ids[node]:
                        bridges.append([neighbor, node])
                else:
                    lowlink[node] = min(lowlink[node], ids[neighbor])


        for i in range(n):
            if i not in visited:
                dfs(i, -1)

        return bridges
        
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