<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>> criticalConnections(int n, vector<vector<int>>& connections) {
        // Create the graph
        vector<vector<int>> graph(n);
        for (const auto& conn : connections) {
            graph[conn[0]].push_back(conn[1]);
            graph[conn[1]].push_back(conn[0]);
        }
        
        // Initialize variables
        vector<int> disc(n, -1), low(n, -1);
        vector<vector<int>> result;
        int time = 0;
        
        // Helper function for DFS
        function<void(int, int)> dfs = [&](int u, int parent) {
            disc[u] = low[u] = time++;
            for (int v : graph[u]) {
                if (v == parent) continue;
                if (disc[v] == -1) { // If v is not visited
                    dfs(v, u);
                    low[u] = min(low[u], low[v]);
                    if (low[v] > disc[u]) { // Check if the edge is a critical connection
                        result.push_back({u, v});
                    }
                } else { // If v is already visited
                    low[u] = min(low[u], disc[v]);
                }
            }
        };
        
        // Start DFS from the first node (assumes the graph is connected)
        dfs(0, -1);
        return result;
    }
};
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