<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>> ans;
    void dfs(vector<vector<int>> &graph , int n , vector<int>&lowest , vector<int> &time , int &timer , vector<int> &visited, int i , int parent ){
        visited[i] = 1;
        time[i] = timer;
        lowest[i] = timer;
        timer++;
        bool skipped = false;
        for(auto j : graph[i]){
            if(j == parent)  continue;
            if(visited[j] == 0 ){
                dfs(graph , n , lowest , time , timer , visited , j , i);
            }
            lowest[i] = min({lowest[i] , lowest[j] });
            if(time[i] < lowest[j] ) ans.push_back({i,j});
        }
    }

    vector<vector<int>> criticalConnections(int n, vector<vector<int>>& connections) {
        vector<int> lowest(n );
        vector<int> time(n);
        int timer = 1;
        vector<vector<int>> graph(n);
        for(auto edge : connections ){
            graph[edge[0]].push_back(edge[1]);
            graph[edge[1]].push_back(edge[0]);
        }
        vector<int> visited(n,0);
        dfs(graph , n , lowest , time , timer , visited , 0 , -1 ) ; 
        return ans;
    }
};
199018