<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]]:
        map_connections = defaultdict(list)
        result=[]
        time = 0
        for i in range(len(connections)):
            map_connections[connections[i][0]].append(connections[i][1])
            map_connections[connections[i][1]].append(connections[i][0])

        
        def critical_connections(timestamp,vist,map_connections,key,prev_key):
            nonlocal time
            vist[key]=1
            timestamp[key]=time
            time=time+1
            curent_time = timestamp[key]

            for i in map_connections[key]:
                if i == prev_key:
                    continue
                if vist[i]==0:
                    critical_connections(timestamp,vist,map_connections,i,key)
                
                timestamp[key] = min(timestamp[key],timestamp[i])
                if curent_time<timestamp[i]:
                    result.append([key,i])

                
        
        vist=[0 for i in range(n+1)]
        timestamp=[0 for j in range(n+1)]
        
        critical_connections(timestamp,vist,map_connections,0,-1)
        
        return result

                



        
        
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