<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:

    # TODO: use union-find to group SCC; see comment - https://leetcode.com/problems/critical-connections-in-a-network/description/comments/1566589
    def criticalConnections(self, n: int, connections: List[List[int]]) -> List[List[int]]:
        graph = [[] for _ in range(n)]
        for a, b in connections:
            graph[a].append(b)
            graph[b].append(a)

        # NOTE: below is an adaptation of Tarjan's strongly connected components algorithm on Wikipedia - https://en.wikipedia.org/wiki/Tarjan%27s_strongly_connected_components_algorithm#The_algorithm_in_pseudocode
        v_to_sid, self.sid = defaultdict(lambda: -1), 0
        stk, on_search_stk = [], set()
        v_idxs, lowlinks, self.idx = defaultdict(lambda: None), defaultdict(lambda: -1), 0
        def _mark_strongly_connected_component(v, parent):
            v_idxs[v] = lowlinks[v] = self.idx
            self.idx += 1
            stk.append(v)
            on_search_stk.add(v)

            for w in graph[v]:
                if w == parent:
                    continue
                elif v_idxs[w] is None:
                    _mark_strongly_connected_component(w, v)
                    lowlinks[v] = min(lowlinks[v], lowlinks[w])
                elif w in on_search_stk:
                    lowlinks[v] = min(lowlinks[v], v_idxs[w])
            
            if v_idxs[v] == lowlinks[v]:
                while stk and stk[-1] != v:
                    w = stk.pop()
                    on_search_stk.remove(w)
                    v_to_sid[w] = self.sid
                if stk and stk[-1] == v:
                    v = stk.pop()
                    on_search_stk.remove(v)
                    v_to_sid[v] = self.sid
                    self.sid += 1

        for v in range(n):
            if v_idxs[v] is None: _mark_strongly_connected_component(v, -1)

        res = []
        for a, b in connections:
            if v_to_sid[a] != v_to_sid[b]: res.append((a, b))

        return res
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