<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 findingbridges(unordered_map<int,vector<int>>& adjList, unordered_map<int,bool> & visited, vector<int> & tin , vector<int> & low, int timer,int src,int parent,vector<vector<int>> & connections){

        visited[src]=true;//we reached here 
        tin[src]=timer;
        low[src]=timer;


        for(auto nbr: adjList[src]){
            if(nbr==parent) continue;
            else if(visited[nbr]){
                //just update the low 
                low[src]=min(low[src],low[nbr]);
            }
            else if(!visited[nbr]){
                //make a dfs call for the ahead edges 
                findingbridges(adjList,visited,tin,low,timer+1,nbr,src,connections);
                low[src]=min(low[src],low[nbr]);
                if(low[nbr]>tin[src]){
                    ///means we got a bridge
                    // low[src]=low[nbr];
                    connections.push_back({src,nbr});
                }
            }
        }
    }
    vector<vector<int>> criticalConnections(int n, vector<vector<int>>& connections) {
        //no of bridges nikalne h 
        //will use tarjan algo 

        //making adjlist
        unordered_map<int,vector<int>> adjList;
        for(auto i: connections){
            adjList[i[0]].push_back(i[1]);
            adjList[i[1]].push_back(i[0]);
        }

        vector<int> tin(n,0);//first insertion time
        vector<int> low(n,0);//updates if gets a second path to be reached
        vector<vector<int>> connectionans;
        unordered_map<int,bool> visited;

        findingbridges(adjList,visited,tin,low,1,0,-1,connectionans);

        return connectionans;

    }
};
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