<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]]:
        bridges = []
        adj_lst = {}
        visited = set()
        discovery_time = [float("inf")] * n
        # low - min id from neighbours ids and itself id
        low = [float("inf")] * n
        parent = [-1] * n
        time = 0
        for x, y in connections:
            adj_lst.setdefault(x, set()).add(y)
            adj_lst.setdefault(y, set()).add(x)

        def dfs(node, visited, parent, low, discovery_time):
            nonlocal time
            visited.add(node)
            discovery_time[node] = low[node] = time
            time += 1

            for nei in adj_lst.get(node, []):
                if nei not in visited:
                    parent[nei] = node
                    dfs(nei, visited, parent, low, discovery_time)

                    low[node] = min(low[nei], low[node])

                    # if nei_node doesn't connect to any node with less discover time than their edge is a bridge
                    if low[nei] > discovery_time[node]:
                        bridges.append([node, nei])

                elif nei != parent[node]:
                    low[node] = min(low[node], discovery_time[nei])

        for node in range(n):
            if node not in visited:
                dfs(node, visited, parent, low, discovery_time)

        return bridges
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