<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: int, connections: List[List[int]]) -> List[List[int]]:

        l= len(connections)
        g=[[] for _ in range(n)]
        for i in range(l):
            g[connections[i][0]].append(connections[i][1])
            g[connections[i][1]].append(connections[i][0])
        print(g)
        vis=[0]*n
        tin=[0]*n
        low=[0]*n
        bridges=[]
        self.timer=0

        return self.dfs(0,-1, vis, g, tin,low,bridges)
    def dfs(self,node, parent, vis, adj, tin, low, bridges):
        # print(bridges, tin, low)
        vis[node]=1
        tin[node], low[node]=self.timer+1, self.timer+1
        self.timer+=1
        for i in adj[node]:
            if i==parent:
                continue
            if vis[i]==0:
                self.dfs(i,node, vis, adj, tin, low, bridges)
                low[node]= min(low[node], low[i])
                if low[i]>tin[node]:
                    print("Sdf")
                    bridges.append([node, i])

            else:
                low[node]=min(low[node], low[i])

        return bridges




            
        
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