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.
Example 1:
Input: n = 4, connections = [[0,1],[1,2],[2,0],[1,3]]
Output: [[1,3]]
Explanation: [[3,1]] is also accepted.
Example 2:
Input: n = 2, connections = [[0,1]]
Output: [[0,1]]
Constraints:
-
2 <= n <= 105 -
n - 1 <= connections.length <= 105 -
0 <= ai, bi <= n - 1 -
ai != bi - There are no repeated connections.
SOLUTION:
from collections import defaultdict
class Solution:
def DFS(self, graph, node, v, parent, disc, low):
v.add(node)
disc[node] = self.t
low[node] = self.t
self.t += 1
for j in graph[node]:
if j not in v:
parent[j] = node
self.DFS(graph, j, v, parent, disc, low)
low[node] = min(low[node], low[j])
elif j != parent[node]:
low[node] = min(low[node], disc[j])
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)
self.t = 0
parent = defaultdict(lambda: -1)
disc = defaultdict(lambda: float('inf'))
low = defaultdict(lambda: float('inf'))
v = set()
for i in range(n):
if i not in v:
self.DFS(graph, i, v, parent, disc, low)
return [[u, v] for u, v in connections if low[u] > disc[v] or low[v] > disc[u]]
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