<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]]:
        adj = {}
        for a, b in connections:
            if a not in adj:
                adj[a] = []
            if b not in adj:
                adj[b] = []
            
            adj[a].append(b)
            adj[b].append(a)
        
        ranks = [-1] * n
        seen = [False] * n
        ans = []
        
        # print(adj)
        
        def helper(curr, parent, k):
            # print(f"{curr} parent={parent}")
            ranks[curr] = k
            seen[curr] = True
            
            lowest_found = None
            
            for child in adj[curr]:
                if child == parent:
                    continue
                if seen[child]:
                    if lowest_found is None or ranks[child] < lowest_found:
                        lowest_found = ranks[child]
                        continue
                    continue
                
                lowest_result = helper(child, curr, k+1)
                if lowest_result is None or ranks[curr] < lowest_result:
                    # print(f"from {curr} to {child} {ranks[curr]} to {lowest_result}")
                    ans.append([curr, child])
                
                if lowest_result is not None:
                    if lowest_found is None or lowest_result < lowest_found:
                        lowest_found = lowest_result
            return lowest_found
        
        helper(0, -1, 0)
        return ans
                    
70988