<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:
    #Tarjan Algo
    #https://takeuforward.org/graph/bridges-in-graph-using-tarjans-algorithm-of-time-in-and-low-time-g-55/
    #https://www.geeksforgeeks.org/bridge-in-a-graph/
    def criticalConnections(self, n: int, connections: List[List[int]]) -> List[List[int]]:
        graph=collections.defaultdict(set)
        for x,y in connections:
            graph[x].add(y)
            graph[y].add(x)
        
        def dfs(u,parent): 
            nonlocal time
            # Mark the current node as visited and print it 
            visited[u]= True

            # Initialize discovery time and low value 
            disc[u] = time
            low[u] = time
            time+=1

            #Recur for all the vertices adjacent to this vertex 
            ans=[]
            for v in graph[u]: 
                # If v is not visited yet, then make it a child of u 
                # in DFS tree and recur for it 
                if not visited[v] : 
                    ans.extend(dfs(v,u))

                    # Check if the subtree rooted with v has a connection to 
                    # one of the ancestors of u 
                    low[u] = min(low[u], low[v]) 
                    ''' If the lowest vertex reachable from subtree 
                    under v is below u in DFS tree, then u-v is 
                    a bridge'''
                    if low[v] > disc[u]: 
                        ans.append([u,v])

                #if already visited and not parent:backedge so can't be bridge, just update low time
                elif v != parent: # Update low value of u for parent function calls. 
                    low[u] = min(low[u], low[v]) 
                #pass when v=parent
            return ans

        visited = [False] * (n) 
        disc = [float("Inf")] * (n) 
        low = [float("Inf")] * (n) 

        # Call the recursive helper function to find bridges 
        # in DFS tree rooted with vertex 'i' 
        ans=[]
        time=0
        for i in range(n): 
            if not visited[i]: 
                ans.extend(dfs(i, -1))
        return ans
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