<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]]:
        vis = [[0,-1] for _ in range(n)]
        tin_tmin = [[0,0] for _ in range(n)]
        adj = [[] for _ in range(n)]
        for con in connections:
            adj[con[0]].append(con[1])
            adj[con[1]].append(con[0])
        # print(degree)
        ans = []
        self.t = -1
        def dfs(s):
            self.t+=1
            tin_tmin[s] = [self.t,self.t]
            vis[s][0] = 1
            for u in adj[s]:
                if(vis[u][0]==0):
                    vis[u][1] = s
                    dfs(u)
                    tin_tmin[s][1] = min(tin_tmin[s][1],tin_tmin[u][1])
                    if(tin_tmin[u][1]>tin_tmin[s][0]):
                        ans.append((s,u))
                elif(vis[u][0]==1 and vis[s][1]!=u):
                    tin_tmin[s][1] = min(tin_tmin[s][1],tin_tmin[u][1])
        dfs(0)
        # print(tin_tmin)
        return ans
                
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