<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>

from collections import defaultdict
class Solution:
    def get_connected_components(self, n, connections):
        visited = [False]*n
        def dfs(node):
            visited[node] = True
            for connection in connections[node]:
                if not visited[connection]:
                    dfs(connection)
        count = 0
        for i in range(n):
            if not visited[i]:
                dfs(i)
                count += 1
        return count
    
    def brute(self, n, connections):
        conns = defaultdict(list)
        for conn in connections:
            conns[conn[0]].append(conn[1])
            conns[conn[1]].append(conn[0])
        ans = []
        for connection in connections:
            conns[connection[0]].remove(connection[1])
            conns[connection[1]].remove(connection[0])
            count = self.get_connected_components(n,conns)
            if count > 1:
                ans.append(connection)
            conns[connection[0]].append(connection[1])
            conns[connection[1]].append(connection[0])
        return ans

    def criticalConnections(self, n: int, connections: List[List[int]]) -> List[List[int]]:
        conns = defaultdict(list)
        for conn in connections:
            conns[conn[0]].append(conn[1])
            conns[conn[1]].append(conn[0])
        self.time = 1
        disc = [0]*n
        low = [0]*n
        visited = [False]*n
        ans = []
        def dfs(node, disc, low, parent):
            visited[node] = True
            disc[node] = self.time
            self.time += 1
            low[node] = disc[node]
            children = 0
            for i in conns[node]:
                if not visited[i]:
                    dfs(i, disc, low, node)
                    low[node] = min(low[node], low[i])
                    children += 1
                    
                    if low[i] > disc[node]:
                        ans.append([i, node])
                elif parent != i:
                    low[node] = min(low[node], disc[i]) 
        dfs(0, disc, low, -1)
        # print(disc)
        # print(low)
        # print(ans)
        return ans
71396