<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>> bridges;
    int timer = 0;
    void dfs(int u, int parent, vector<int> &visited, vector<int> &in_time, vector<int>& low_time, vector<vector<int>> &edges){
        visited[u] = 1;
        in_time[u] = low_time[u] = timer++;
        for(auto v : edges[u]){
            if(v == parent) continue;
            if(visited[v]){
                low_time[u] = min(low_time[u], in_time[v]);
            }else{
                dfs(v, u, visited, in_time, low_time, edges);
                low_time[u] = min(low_time[u], low_time[v]);
                if(low_time[v] > in_time[u]){
                    bridges.push_back({u, v});
                }
            }
        }
    }
    vector<vector<int>> criticalConnections(int n,
                                            vector<vector<int>>& connections) {
        vector<int> visited(n, 0);
        vector<int> in_time(n, -1);
        vector<int> low_time(n, -1);
        vector<vector<int>> edges(n);

        for(vector<int> conn : connections){
            edges[conn[0]].push_back(conn[1]);
            edges[conn[1]].push_back(conn[0]);
        }
        
        for(int i=0; i<n; i++){
            if(!visited[i]){
                dfs(i, -1, visited, in_time, low_time, edges);
            }
        }

        return bridges;
    }
};
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