<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:
    int timer=1;
    
    void dfs(int node, int parent, vector<bool>&vis,map<int,vector<int>>&adj, int tin[] , int low[],vector<vector<int>>&bridges){
        vis[node]=true;
        tin[node]=low[node]=timer;
        timer++;

        for(auto it : adj[node]){
            if(it==parent)continue;
            if(!vis[it]){
                //if not visited we will go to neighbour node
                dfs(it, node, vis, adj, tin, low,bridges);

                //update low when coming back
                low[node]=min(low[node],low[it]);

                //if low of neighbour is more then the time of insertion of node then it is a bridge
                if(low[it]>tin[node])bridges.push_back({it,node});
            }
            else{
                low[node]=min(low[node],low[it]);
            }
        }
    }

    vector<vector<int>> criticalConnections(int n, vector<vector<int>>& connections) {
        map<int,vector<int>>adj;
        for(auto  it:connections){
            adj[it[0]].push_back(it[1]);
            adj[it[1]].push_back(it[0]);
        }
        //visisted
        vector<bool>vis(n, false);
        //time of insertion
        int tin[n];
        //lowest time
        int low[n];

        vector<vector<int>>bridges;
        dfs(0, -1, vis, adj, tin , low,bridges);
        return bridges;
        
    }
};
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