<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]]:
        adjMap = { i : [] for i in range(n)}
        ansCon = set((min(n1,n2), max(n1,n2)) for (n1,n2) in connections)
        for n1, n2 in connections:
            adjMap[n1].append(n2)
            adjMap[n2].append(n1)

        rank = [-2] * n

       

        curRank = 0
        def dfs(node, curRank):
            if rank[node] != -2:
                return rank[node]
            else:
                rank[node] = curRank
                curMinRank = curRank
                for adj in adjMap[node]:
                    #parent check: -> parent will have rank -1
                    if rank[adj] != curRank - 1:
                        minRankFound = dfs(adj, curRank + 1)
                        if minRankFound <= curRank:# means that this node will lead to a cycle
                             ansCon.remove((min(adj, node), max(adj, node)))
                        curMinRank = min(curMinRank, minRankFound)
                return curMinRank
                            
        dfs(0,0)

        return [[n1,n2] for (n1,n2) in ansCon]
                #forget about parent !!! -> when visit 3 -> remove 1
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