<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:
    rank  = {}
    graph = defaultdict(list)
    conn_dict = {}

    def criticalConnections(self, n: int, connections: List[List[int]]) -> List[List[int]]:
        self.formGraph(n,connections)
        self.dfs(0,0)
        ans = []
        for u,v in self.conn_dict:
            ans.append([u,v])
        return ans

    def formGraph(self,n,connections):
        self.rank  = {}
        self.graph = defaultdict(list)
        self.conn_dict = {}

        for i in range(n):
            self.rank[i] = None

        for u,v in connections:
            self.graph[u].append(v)
            self.graph[v].append(u)
            self.conn_dict[(min(u,v),max(u,v))] = True

    def dfs(self,node,rank):
        if self.rank[node]:
            return self.rank[node]
        self.rank[node] = rank
        min_rank = rank + 1
        for neighbour in self.graph[node]:
            if self.rank[neighbour] and self.rank[neighbour] == rank-1:
                continue
            
            recursive_rank = self.dfs(neighbour,rank+1)
            if recursive_rank <= rank:
                del self.conn_dict[(min(node,neighbour),max(node,neighbour))]
            min_rank = min(recursive_rank,min_rank)

        return min_rank


        
        
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