<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>

'''
an edge is critical if there is only one way of getting from the node to the other node
if keep track of two times: 
    1. minimum time to reach a node
    2. time it takes to reach a node
    
0 -> 1, 2
1 -> 0, 2, 3
2 -> 0, 1
3 -> 1

time = 0, 0, 0, 2
dist = 0, 1, 2, 2
'''
class Solution:
    def criticalConnections(self, n: int, connections: List[List[int]]) -> List[List[int]]:
        graph = [[] for _ in range(n)]
        result = []
        
        for a, b in connections:
            graph[a].append(b)
            graph[b].append(a)
            
        reach = [-1] * n
        dist  = [-1] * n

        def dfs(node, parent, time):
            reach[node] = time
            dist[node]  = time
            
            for neigh in graph[node]:
                if reach[neigh] == -1:
                    dfs(neigh, node, time + 1)
                    
                    if reach[neigh] > dist[node]:
                        result.append([node, neigh])
                    else:
                        reach[node] = min(reach[node], reach[neigh])
                
                elif neigh != parent:
                    reach[node] = min(reach[node], reach[neigh])
                    
        
        dfs(0, -1, 0)
        return result
        
        
        
        
        
        
        
        
        
                    
                    
                    
                    
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