<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;  // Timer to keep track of discovery times
private:
    void dfs(int node, int parent, vector<int> &visited, vector<int> adj[], 
             vector<int> &tin, vector<int> &low, vector<vector<int>> &bridges) {
        visited[node] = 1;
        tin[node] = low[node] = timer++;

        for (int neighbor : adj[node]) {
            if (neighbor == parent) continue;
            if (!visited[neighbor]) {
                dfs(neighbor, node, visited, adj, tin, low, bridges);
                low[node] = min(low[node], low[neighbor]);

                // Check if the edge is a bridge
                if (low[neighbor] > tin[node]) {
                    bridges.push_back({node, neighbor});
                }
            } else {
                low[node] = min(low[node], tin[neighbor]);
            }
        }
    }

public:
    vector<vector<int>> criticalConnections(int n, vector<vector<int>>& connections) {
        timer = 0;
        vector<int> adj[n];
        adj->reserve(n);  // Reserve space for adjacency list to optimize memory

        // Build adjacency list
        for (const auto &connection : connections) {
            adj[connection[0]].push_back(connection[1]);
            adj[connection[1]].push_back(connection[0]);
        }

        vector<int> visited(n, 0), tin(n, -1), low(n, -1);
        vector<vector<int>> bridges;
        bridges.reserve(connections.size());  // Reserve space for potential bridges

        // Start DFS from node 0 (assuming graph is connected)
        dfs(0, -1, visited, adj, tin, low, bridges);

        return bridges;
    }
};

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