<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 __init__(self):
        self.timer = 1
    def dfs(self, adj, vis, node, parent, time_in, lowtime, res):
        vis[node] = 1
        time_in[node], lowtime[node] = self.timer, self.timer
        self.timer += 1
        for nei in adj[node]:
            # Do not take parent element of a ndoe into consideration.
            if nei == parent:
                continue
            # Visit the node.
            if vis[nei] == 0:
                self.dfs(adj, vis, nei, node, time_in, lowtime, res)
                # Take the lowest time of all neighbouring nodes to reach the nei node.
                lowtime[node] = min(lowtime[node], lowtime[nei])
                # If the nei node only reachable via node, then its a bridge.
                if lowtime[nei] > time_in[node]: 
                    print(node, nei)
                    res.append([node, nei])
            else:
                # Take the lowest time of all neighbouring nodes to reach the nei node.
                lowtime[node] = min(lowtime[node], lowtime[nei])

    def criticalConnections(self, n: int, connections: List[List[int]]) -> List[List[int]]:
        adj = [[] for i in range(n)]
        for i, j in connections:
            adj[i].append(j)
            adj[j].append(i)
        vis = [0 for i in range(n)]
        time_in = [0 for i in range(n)]
        lowtime = [0 for i in range(n)]
        res = []
        self.dfs(adj, vis, 0, -1, time_in, lowtime, res)
        return res
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