<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, connections):
        used, tin, fup = [0]*n, [-1]*n, [-1]*n
        self.timer, ans = 0, []
        graph = defaultdict(list)
        
        def dfs(node, par = -1):
            used[node] = 1
            tin[node] = fup[node] = self.timer + 1
            self.timer += 1
            for child in graph[node]:
                if child == par: continue
                if used[child] == 1:
                    fup[node] = min(fup[node], tin[child])
                else:
                    dfs(child, node)
                    fup[node] = min(fup[node], fup[child])
                    if fup[child] > tin[node]: ans.append([node, child])
        
        for i, j in connections:
            graph[i].append(j)
            graph[j].append(i)
            
        for i in range(n):
            if not used[i]: dfs(i)
                
        return ans
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