<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]]:
        graph = {i: [] for i in range(n)}
        rank = [-2] * n
        conn_dict = {}
       
        
        for node1, node2 in connections:
            graph[node1].append(node2)
            graph[node2].append(node1)
            conn_dict[(min(node1, node2), max(node1, node2))] = 1
        # print("graph", graph)

        def dfs(node: int, r: int) -> int:
            # print("node", node)
            # print("rank", rank)
            if rank[node] != -2:
                return rank[node]
            
            rank[node] = r
            min_rank = r + 1
            for neighbor in graph[node]:
                if rank[neighbor] == r-1:
                    continue
                
                recursive_rank = dfs(neighbor, r + 1)
                if recursive_rank <= r:
                    conn_dict[(min(node, neighbor), max(node, neighbor))] = 0
                    # print("conn_dict", conn_dict)
                min_rank = min(min_rank, recursive_rank)
            return min_rank
        dfs(0,0)
        res = []
        # print(conn_dict)
        for node1,node2 in conn_dict:
            if conn_dict[(node1, node2)] == 1:
                res.append([node1, node2])
        return res
                       
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