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Critical Connections in a Network -Leetcode 1192

Sameeksha Bhatia
An IT undergrad interested in Python,Django and DSA.
・2 min read

1. Problem Statement

There are n servers numbered from 0 to n - 1 connected by undirected server-to-server connections forming a network where connections[i] = [ai, bi] represents a connection between servers ai and bi. Any server can reach other servers directly or indirectly through the network.

A critical connection is a connection that, if removed, will make some servers unable to reach some other server.

Return all critical connections in the network in any order.

2. Lets try to understand the approach

disc = discovery time of each node in the graph
|The concept of 'LOW'|
low values means to which scc this node belongs to and that if this node is reachable to
its ancestor (basically is there any way , if yes then we assign the low value of this node to the low value of its ancestor which will signify that this node belongs to the SCC and that the edge from this node to the ancestor node is the back edge
when this node will backtrack to its parent node then it will say that look the connection between you(Parent) and me(child) is not a critical connection because I can be dicovered without your help(Parent) and we all belong to one SCC hence we update the parents low value to the min of low[parent],low[child])
Next imp thing :
|Condition to check if this edge is a bridge |

    eg : SCC1 -----> SCC2
        disc=0       disc =1
        low=0        low=1
    SCC1 =      1->2->3->1           SCC2= 4->5->4
         disc   0  1  2  0           disc   3  4  3
         low    0  0  0  0           low    3  3  3

    suppose there s an edge from 1 to 4 and we start our dfs from node 1 then we know that 
    the discovery of node 1 is done before the discovery of node 4 
    now we cant simply compare the dicovery time of the nodes(thats not correct)
    hence we use the low value of the chid to check 
    | if dicovery[parent] < low[child] |
    MEANS , That the dicovery ofthe parent was done before that of child hence its value is small if the edge is critical edge 
    else if this edge is not a critical edge then the low value of the child is lesser than or equal to parent
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def criticalConnections(self, n: int, connections: List[List[int]]) -> List[List[int]]:

        disc = [-1]*n
        low = [-1]*n
        self.time = 0
        graph = defaultdict(list)
        bridges = []
        for u,v in connections:

        def dfs(node,parent):
            if disc[node] != -1:

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

            for neighbor in graph[node]:
                if neighbor == parent:

                low[node] = min(low[neighbor],low[node])

                if disc[node] < low[neighbor] :

        return bridges
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Time Complexity: O(V+E)

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