<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:
    void dfs(vector<int> adj[], unordered_map<int, bool>& visited, int index, int parent, int time, 
    vector<int>& insertion, vector<int>& low, vector<vector<int>>& ans){
        visited[index] = true;
        low[index] = time, insertion[index] = time;
        time++;
        for(auto x: adj[index]){
            if(visited.find(x) == visited.end()){
                //if this current node has not been visited, we visit it
                dfs(adj, visited, x, index, time+1, insertion, low, ans);
                low[index] = min(low[index], low[x]);
                if(low[x] > insertion[index]){
                    //this is a bridge
                    ans.push_back({index, x});
                }
            }
            else{
                //if this node has been visited, we check if the node is the parent or not
                if(x == parent) continue;
                else{
                    //now we check if the particular edge is a bridge or not
                    low[index] = min(low[index], low[x]);
                }
            }
        }
    }
    vector<vector<int>> criticalConnections(int n, vector<vector<int>>& connections) {
        //we make the adjacency list of the connections
        ios_base::sync_with_stdio(false);
        cin.tie(nullptr); cout.tie(nullptr);
        vector<int> adj[n];
        for(int i=0; i<=connections.size()-1; i++){
            int u = connections[i][0];
            int v = connections[i][1];
            //undirected graph and so we push into both the lists
            adj[u].push_back(v);
            adj[v].push_back(u);
        }
        //simply find bridges in a graph
        vector<int> insertion(n, -1);
        vector<int> low(n, -1);
        unordered_map<int, bool> visited;
        vector<vector<int>> ans;
        int time = 0;
        //we have a connected graph
        //mark the node as visited at the time of the call
        dfs(adj, visited, 0, -1, time, insertion, low, ans);
        return ans;
    }
};
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