<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 dfs(self,adj,par,e,vis,lowtime,timein,i,ans):
        vis[e]=1
        timein[e]=i
        lowtime[e]=i
        i+=1
        for j in range(len(adj[e])):
            if vis[adj[e][j]]==0:
                self.dfs(adj,e,adj[e][j],vis,lowtime,timein,i,ans)
                if timein[e]<lowtime[adj[e][j]]:
                    ans.append([e,adj[e][j]])
                lowtime[e]=min(lowtime[adj[e][j]],lowtime[e])
            elif adj[e][j]!=par:
                lowtime[e]=min(lowtime[e],lowtime[adj[e][j]])
        return 


    def criticalConnections(self, n: int, connections: List[List[int]]) -> List[List[int]]:
        adj=[[] for i in range(n)]
        e=len(connections)
        for i in range(e):
            adj[connections[i][0]].append(connections[i][1])
            adj[connections[i][1]].append(connections[i][0])
        ans=[]
        # par=[0 for i in range(n)]
        vis=[0 for i in range(n)]
        lowtime=[1 for i in range(n)]
        timein=[1 for i in range(n)]
        vis[0]=1
        self.dfs(adj,-1,0,vis,lowtime,timein,1,ans)
        return ans

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