<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 tarjan_dfs(const vector<vector<int>>& graph, int v, int parent,
                    int& time, vector<int>& disc, vector<int>& low,
                    vector<bool>& visited, vector<pair<int, int>>& bridges) {
        visited[v] = true;
        disc[v] = low[v] = time++;
        bool parent_ignored = false;

        for (const auto& to : graph[v]) {
            if (to == parent && !parent_ignored) {
                parent_ignored = true; // Ignore the parent edge

                // This is only useful for multiply connected graphs
                continue;
            }

            if (visited[to]) {
                // Update low value of v for parent function calls
                low[v] = min(low[v], disc[to]);
            } else {
                tarjan_dfs(graph, to, v, time, disc, low, visited, bridges);
                low[v] = min(low[v], low[to]);

                // If the lowest vertex reachable from subtree under to is
                // below v in DFS tree, then v-to is a bridge
                if (low[to] > disc[v]) {
                    bridges.push_back({v, to});
                }
            }
        }
    }

    // Function to find critical connections in a network
    // In graph theory, this is called a *bridge*. We can use Tarjan's algorithm
    // to find all bridges in a graph.
    vector<vector<int>>
    criticalConnectionsPair(int n, vector<pair<int, int>>& connections) {
        // Build the lookup adjacency list
        vector<vector<int>> graph(n);
        for (const auto& [u, v] : connections) {
            graph[u].push_back(v);
            graph[v].push_back(u);
        }

        // Initialize variables for Tarjan's algorithm
        // We need to find the DFS time and the lowest DFS time reachable for
        // each vertex
        vector<int> disc(n, -1), low(n, -1);
        vector<bool> visited(n, false);
        vector<pair<int, int>> bridges;
        int time = 0;

        // Call the DFS function for each vertex
        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                tarjan_dfs(graph, i, -1, time, disc, low, visited, bridges);
            }
        }

        // Convert and deduplicate the bridges into the required output format

        vector<vector<int>> result;
        for (const auto& [u, v] : bridges) {
            result.push_back({u, v});
        }

        return result;
    }

    // Wrapper for vector<vector<int>> input
    vector<vector<int>> criticalConnections(int n,
                                            vector<vector<int>>& connections) {
        vector<pair<int, int>> edges(connections.size());

        for (const auto& connection : connections) {
            edges.emplace_back(connection[0], connection[1]);
        }
        return criticalConnectionsPair(n, edges);
    }
};
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