<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]]:

        graph = defaultdict(list)

        for u, v in connections:
            graph[u].append(v)
            graph[v].append(u)

        ids = [-1] * n      # Discovery times of nodes
        low = [0] * n       # Lowest point reachable
        stack = []          # To store edges
        current_id = 0      # Global discovery time
        bccs = []           # List of biconnected components

        def dfs(at, parent):
            nonlocal current_id
            ids[at] = low[at] = current_id
            current_id += 1
            children = 0

            for to in graph[at]:
                if to == parent:
                    continue

                if ids[to] == -1:  # Node `to` has not been visited
                    stack.append((at, to))  # Store edge in the stack
                    children += 1
                    dfs(to, at)
                    low[at] = min(low[at], low[to])

                    # If `low[to] >= ids[at]`, then `at` is an articulation point
                    if low[to] >= ids[at]:
                        component = []
                        while stack and stack[-1] != (at, to):
                            component.append(stack.pop())
                        component.append(stack.pop())  # Include the edge (at, to)
                        bccs.append(component)

                elif ids[to] < ids[at]:  # Back edge
                    stack.append((at, to))
                    low[at] = min(low[at], ids[to])

        for i in range(n):
            if ids[i] == -1:
                dfs(i, -1)

                # If there are remaining edges in the stack after DFS, it's a BCC
                if stack:
                    component = []
                    while stack:
                        component.append(stack.pop())
                    bccs.append(component)
        
        res = []
        for arr in bccs:
            if len(arr) == 1:
                res.append([arr[0][0], arr[0][1]])
        return res
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