<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 = 1;

void dfs(int node, int parent,vector<bool> &visited, 
        vector<int> adj[], vector<int> &in_time , 
        vector<int> &low_time, vector<vector<int>> &ans ){
            visited[node] = true;
            in_time[node] = timer;
            low_time[node] = timer;
            timer++;

            for(auto neighbour:adj[node]){
                if(neighbour == parent) 
                    continue;
                if(!visited[neighbour]){
                    dfs(neighbour, node, visited, adj, in_time, low_time, ans);
                    low_time[node] = min(low_time[node], low_time[neighbour]);

                    if(low_time[neighbour] > in_time[node])
                        ans.push_back({node, neighbour}); // this is critical bridge
                    
                }
                else
                    low_time[node] = min(low_time[node], low_time[neighbour]);

            }
        }




public:
    vector<vector<int>> criticalConnections(int n, vector<vector<int>>& connections) {
        // making graph

        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<bool> visited(n, false);
        vector<int> in_time(n);
        vector<int> low_time(n);
        vector<vector<int>> ans;
        int node = 0;
        int parent = -1;

        dfs(node, parent, visited, adj, in_time, low_time, ans);

        return ans;
        
    }
};
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