<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 dfs(self, node, adj, time, low, res, parent, curr_time):
        curr_time[0] += 1
        for neigh in adj[node]:
            if time[neigh] == -1:
                time[neigh] = low[neigh] = curr_time[0]
                self.dfs(neigh, adj, time, low, res, node, curr_time)
                low[node] = min(low[neigh], low[node])

                if low[neigh] > time[node]:
                    res.append([node, neigh])

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

    def criticalConnections(self, n: int, connections: List[List[int]]) -> List[List[int]]:
        time = [-1] * n
        low = [-1] * n
        res = []
        time[0], low[0] = 0, 0
        curr_time = [0]
        adj = [[] for _ in range(n)]
        for i, j in connections:
            adj[i].append(j)
            adj[j].append(i)

        self.dfs(0, adj, time, low, res, -1, curr_time)

        print(f'time->{time} low->{low}')
        return res
64448