<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, edges: List[List[int]]) -> List[List[int]]:
        adj = [[] for i in range(n)]
        for edge in edges:
            u_, v_ = edge
            adj[u_].append(v_)
            adj[v_].append(u_)
        
        visited = {i:0 for i in range(n)}; criticaledges = set()
        time = 1
        firstTime = dict()

        def dfs(node, parent, fC):
            # Lets mark the node as visited
            nonlocal time
            #print(node, fC)
            visited[node] = 1
            firstTime[node] = actualdist = fC
            adjnodes = adj[node]
            for nextnode in adjnodes:
                if nextnode == parent:
                    continue
                elif visited[nextnode] == 1:
                    actualdist = min(actualdist, firstTime[nextnode])
                    #print(node, nextnode, actualdist)
                    # This edge is not a critical edge
                else:
                    time += 1
                    actual = dfs(nextnode, node, time)
                    #print(actual > fC, actual, fC, node, nextnode)
                    if actual > fC:
                        criticaledges.add((node, nextnode))
                    else:
                        actualdist = min(actualdist, actual)
            
            print("Final ",node, actualdist)
            return actualdist
        
        dfs(0, None, time)

        #print(criticaledges, time)
        return criticaledges
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