<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<int> visited(n,0);
        vector<int> first_see(n,0);
        vector<int> min_see(n,1e9);
        vector<vector<int>> adj(n);
        for(auto it : connections){
            int u=it[0],v=it[1];
            adj[u].push_back(v);
            adj[v].push_back(u);
        }
        vector<vector<int>> ans;
        int time=0;
        function<void (int parent,int node)> dfs= [&] (int parent,int node){
            time++;
            first_see[node]=time;min_see[node]=time;
            for(auto i : adj[node]){
                if(i==parent){
                    continue;
                }
                if(!visited[i]){
                    visited[i]=1;
                    dfs(node,i);
                    min_see[node]=min(min_see[node],min_see[i]);
                }else{
                    min_see[node]=min(min_see[node],min_see[i]);
                }
                if(first_see[node]<min_see[i]){
                    ans.push_back({node,i});
                }
            }
        };
        for(int i=0;i<n;++i){
            if(!visited[i]){
                visited[i]=1;
                dfs(-1,i);
            }
        }
        return ans;
    }
};
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