<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:
    vector<vector<int>> criticalConnections(int n, vector<vector<int>>& connections) {
        vector<vector<int>>adj(n);
        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);
        }
        vector<int>vis(n,0); // will store time to reach the node;
        vector<int>mintime(n,0); //will store the minimum time to reach a node
        int time=1;
        vector<vector<int>>ans;
        for(int i=0;i<n;i++){
            if(!vis[i]) dfs(i,-1,adj,vis,mintime,time,ans);
        }
        
        return ans;

    }
    

    void dfs(int node, int par, vector<vector<int>>&adj, vector<int>&vis, vector<int>&mintime, int &time, vector<vector<int>>&ans){
        vis[node]=time;
        mintime[node]=time;
        time++;
        for(auto &child:adj[node]){
            if(par==child) continue;
            if(!vis[child]){
                dfs(child,node,adj,vis,mintime,time,ans);
            }
            if(mintime[child]>vis[node]) ans.push_back({node,child});
            mintime[node]=min(mintime[node],mintime[child]);
        }
    }
};
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