<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>

from collections import defaultdict

class Solution:
   def criticalConnections(self, n: int, connections: List[List[int]]) -> List[List[int]]:
    #Kosaraju's Algorithm
    
    #build graph
    graph = {}
    for i in range(n): graph[i] = set()
    for u,v in connections:
        graph[u].add(v)
        graph[v].add(u)
    
    #traverse to get order
    order = []
    visited = [0 for _ in range(n)]
    parents = [-1 for _ in range(n)]
    def get_order(node):
        visited[node] = 1
        for neighbour in graph[node]:
            if not visited[neighbour]:
                parents[neighbour] = node
                get_order(neighbour)
        order.append(node)
    
    #outer loop DFS
    for node in range(n):
        if not visited[node]:
            get_order(node)
    
    #transpose graph
    #however, since this is already undirected graph, just need to remove
    #one out ouf the bidirectional. Just remove the connection from parents to itself
    for node in range(n):
        if parents[node]!=-1:
            graph[parents[node]].remove(node)
    
    #process the scc in decreasing order of finish time
    visited = [0 for _ in range(n)]
    def find_scc(node):
        visited[node] = 1
        for neighbour in graph[node]:
            if not visited[neighbour]:
                find_scc(neighbour)
    
    #outer loop dfs
    critical_edges = []
    while order:
        node = order.pop()
        #new scc if found when the node has not been visited in previous traversal
        if not visited[node]:
            if parents[node]!=-1: 
                critical_edges.append([parents[node], node])
            find_scc(node)
    
    return critical_edges
            
    
        
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