<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>

# Space complexity 1 visited means number of Vertices + 2n means number of 2edges = 2E+V
# time complexity of DFS V+2E
class Solution:
    def criticalConnections(self, n: int, connections: List[List[int]]) -> List[List[int]]:
        adj=defaultdict(list)
        for u,v in connections:
            adj[u].append(v)
            adj[v].append(u)

        time = [float('inf')] * n   
        low = [float('inf')] * n   
        bridges = [] 
        timer = 0   
        seen = set()  

     
        def dfs(node, parent):
            nonlocal timer
            seen.add(node)
            time[node] = low[node] = timer  # Set discovery time and low value
            timer += 1
            
            for neighbor in adj[node]:
                if neighbor == parent:  # Ignore the edge leading back to the parent
                    continue
                if neighbor not in seen:
                    dfs(neighbor, node)
                    # After visiting the neighbor, update the low value for the current node
                    low[node] = min(low[node], low[neighbor])
                    
                    # If the lowest reachable node from neighbor is greater than the discovery time of node,
                    # this is a critical connection (bridge)
                    if low[neighbor] > time[node]:
                        bridges.append([node, neighbor])
                else:
                    # If the neighbor is already visited, update the low value for the current node
                    low[node] = min(low[node], time[neighbor])

        # Perform DFS starting from the first node (node 0)
        dfs(0,-1)
        
        return bridges
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