<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]]:
        graph = defaultdict(list)

        for a,b in connections:
            graph[a].append(b)
            graph[b].append(a)

        def dfs():
            stk = [(0 , 0 , -1)]
            seen = [0 for _ in  range(n)]

            disc = {}
            low = {}

            time = 0

            ans = []

            while stk:
                
                node,state,par = stk.pop()

                if state:
                    for nbr in graph[node]:
                        if nbr != par and disc[node] < low[nbr]:
                            ans.append([node , nbr])
                
                        if nbr != par:
                            low[node] = min(low[node] , low[nbr])
                            # print(node , "nbr" , nbr  , low[nbr], "par" , par)
                        # print(node , low[node])
                    continue
                
                if seen[node]:
                    continue

                disc[node] = time
                low[node] = time
                time += 1

                seen[node] = 1
                stk.append((node , 1 , par))
                for nbr in graph[node]:
                    if not seen[nbr]:
                        stk.append((nbr , 0 , node))
            
            # print(low)
            return ans
        


        return dfs()






        
81615