<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:
// to find bridge

    int time;
    vector<vector<int>> ans;

    void dfs(int u, vector<int> edge[], int disc[], int low[], int parent[]){
        disc[u]=low[u]=time++;

        for(int i=0;i<edge[u].size();i++){
            int v = edge[u][i];
            if(disc[v]==-1){ // not discovered yet
            
                parent[v]=u;
                
                dfs(v,edge,disc,low,parent);

                if(disc[u]<low[v]){ // mean u always dicvoer before low of v is articulation point
                    vector<int> temp;
                    temp.push_back(u);
                    temp.push_back(v);
                    ans.push_back(temp);
                }

                low[u] = min(low[u],low[v]);
                
            }else{
                if(v != parent[u]){
                    low[u] = min(low[u],disc[v]);
                }
            }
        }
    }

    vector<vector<int>> criticalConnections(int n, vector<vector<int>>& connections) {

        time = 1;

        int disc[n];
        int low[n];
        int parent[n];
        for(int i=0;i<n;i++){
            disc[i]=-1;
            low[i]=-1;
            parent[i]=-1;
        }

        vector<int> edge[n];

        for(int i=0;i<connections.size();i++){
            edge[connections[i][0]].push_back(connections[i][1]);
            edge[connections[i][1]].push_back(connections[i][0]);
        }



        for(int i=0;i<n;i++){
            if(disc[i]==-1) dfs(i,edge,disc,low,parent);
        }

        return ans;
        
    }
};
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