<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:
    def criticalConnections(self, n: int, connections: List[List[int]]) -> List[List[int]]:
        """
        An implementation of the criticalConnections function with Tarjan's algorithm.
        """
        # if n <= 2:
        #     return connections

        edges = defaultdict(list)                           # Initialize the edges dictionary
        ranks = [-1] * n                                    # Initialize the ranks list

        # print("Connections: ", connections)

        for edge in connections:                            # Iterate through the connections
            # edge = sorted(edge)                             # Sort the edge to avoid duplicates
            edges[edge[0]].append(edge[1])                  # Add the destination to the source key
            edges[edge[1]].append(edge[0])                 # Add the source to the destination key

        connections = set(map(tuple, map(sorted, connections))) # This will store our critical edges (faster than using a list)
        
        def dfs(rank, node, parent):
            if ranks[node] >= 0:                            # If the rank of the node is greater than or equal to 0
                return ranks[node]                          # The node has been visited before, return the rank
            ranks[node] = rank                              # Otherwise, set the rank of the node to the current rank
            min_rank = rank                                 # Initialize the minimum rank to the current rank
            for neighbor in edges[node]:                    # Iterate through the neighbors of the node
                if neighbor == parent:                      # If the rank of the neighbor is equal to the previous rank (parent)
                    continue                                # Skip the neighbor
                recursive_rank = dfs(rank + 1, neighbor, node)              # Recurse
                if recursive_rank <= rank:                  # If the recursive rank is less than or equal to the current rank (it is a cycle)
                    connections.discard(tuple(sorted((node, neighbor))))    # Remove the edge from the connections set
                min_rank = min(min_rank, recursive_rank)
            return min_rank                                 # Return whatever rank is smaller
        
        dfs(0, 0, -1)                                           # Start at rank 0 and node 0 (there will always be at least one node)

        return connections
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