<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]]:
        ans = []
        graph = [set() for _ in range(n)]
        for edge in connections:
            graph[edge[0]].add(edge[1])
            graph[edge[1]].add(edge[0])
        

        visited = set()
        connected = set()
        level = {}
        def dfs(curr, path, l, prev):
            nonlocal visited, connected, level
            #print("curr ", curr, " con ", connected, " lebels ", level)
            if curr in level:
                return level[curr]
            if curr in visited:
                return float('inf')

            visited.add(curr)

            level[curr] = l
            n = float('inf')
            for child in graph[curr]:
                if child == prev:
                    continue
                val = dfs(child, path, l+1, curr)
                #print("par ", curr, " ch ", child, " val ", val)
                if val != float('inf'):
                    connected.add((curr, child))
                n = min(val, n)
            if n == level[curr]:
                n = float('inf')
            del level[curr]
            return n
            




            

        dfs(0, {0}, 0, -1)
        for edge in connections:
            if (edge[0], edge[1]) in connected or (edge[1], edge[0]) in connected:
                continue
            ans.append(edge)
        #print(connected)
        return ans

78754