<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]]:
        self.rank = dict()
        self.connections_dict = dict()
        self.graph = defaultdict(list)

        self.formGraph(n, connections)

        self.dfs(0, 0) # start at node index 0, with discovery_rank 0


        # left over undeleted connections are the critical connections
        return [[u, v] for u, v in self.connections_dict]
    
    def formGraph(self, n, connections):
        for idx in range(n):
            self.rank[idx] = None
        for u, v in connections:
            self.connections_dict[min(u, v), max(u, v)] = 1
            self.graph[u].append(v)
            self.graph[v].append(u)


    def dfs(self, node, discovery_rank):
        if self.rank[node]:
            return self.rank[node]
        
        self.rank[node] = discovery_rank

        min_rank = discovery_rank + 1

        for neighbor in self.graph[node]:
            if self.rank[neighbor] and self.rank[neighbor] == discovery_rank - 1: # immediate parent
                continue
            
            recursive_rank = self.dfs(neighbor, discovery_rank + 1)
            if recursive_rank <= discovery_rank:  # the recursively returned rank is less than current traversal rank.
                del self.connections_dict[min(node, neighbor), max(node, neighbor)]
            
            min_rank = min(min_rank, recursive_rank)
        
        return min_rank
        
        
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