<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 {
    int timer = 0;
public:
    vector<vector<int>> criticalConnections(int n, vector<vector<int>>& conns) {
        vector<int> gg[n];
        for(auto& it: conns){
            int u = it[0];
            int v = it[1];
            gg[u].push_back(v);
            gg[v].push_back(u);
        }
        vector<int> time(n), ltime(n), vis(n, 0);
        vector<vector<int>> res;
        function<void(int, int)> dfs = [&](int src, int par){
            vis[src] = 1;
            time[src] = ltime[src] = timer;
            timer++;
            for(auto ch: gg[src]){
                if(ch == par) continue;
                if(!vis[ch]){
                    dfs(ch, src);
                    ltime[src] = min(ltime[src], ltime[ch]);
                    if(ltime[ch] > time[src]) res.push_back({src, ch});
                } else ltime[src] = min(ltime[src], ltime[ch]);
            }
        };
        dfs(0, -1);
        return res;
    }
};
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