<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 {
public:
    
    void dfs(int node, int parent,int &timer,vector<int> &disc ,vector<int> &low ,vector<bool> &visited,
               unordered_map<int,list<int>> &adj , vector<vector<int>>& connections ) 
    {
        visited[node] = true ; 
        disc[node] = low[node] = timer++ ; 
    
        //teraverse neighbours/Adjacent 
        for( auto nbr : adj[node] )
        {
            if( nbr == parent ) continue ; 
            
            if( !visited[nbr] )
            {
                dfs( nbr , node ,timer , disc , low , visited, adj, connections ) ;
                //while returning from the call update Low[node]
                low[node] = min( low[node] , low[nbr] ) ;
                
                //check is curr Edge is bridge/crittical 
                if( low[nbr] > disc[node] )
                {
                    //store the nodes in ans
                    connections.push_back( {node,nbr} ) ;
                }
            }
            else{
                //if nbr is visited && not parent means curr edge is 
                //BACK EDGE
                low[node] = min( low[node] , low[nbr] ) ; 
            }
        }
    
    }
    
    
    
    //BRIDEG IN A GRAPH//
    vector<vector<int>> criticalConnections(int n, vector<vector<int>>& connections) {
        
        //Create Adj 
        unordered_map<int,list<int>> adj ; 
        for(int i=0 ; i<connections.size() ; i++)
        {
            int u = connections[i][0] ; 
            int v = connections[i][1] ;
            
            adj[u].push_back(v) ; 
            adj[v].push_back(u) ; 
        }
        
        connections.clear() ;   //we will use this vector to store our ans 
        
        //required
        int timer = 0 ; 
        int parent = 0 ; 
        vector<int> disc(n , -1) ;  //discovery time for any node 
        vector<int> low(n , -1) ;  //LOW : Earlist Possible Time--> of any Node
        vector<bool> visited(n , false) ;   //visited
        
        for(int i=0 ; i<n ; i++)
        {
            if( !visited[i] )
                dfs( i , parent , timer , disc , low , visited, adj , connections ) ; 
        }
        
        return connections ;    //bcz we used this vector to store ANS 
    }
};
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