Created
February 23, 2020 04:24
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from collections import deque | |
class Solution: | |
def bfsNodes(n: int, adj_list: List[List[int]]) -> int: | |
if len(adj_list[0]) == 0: | |
return 0 | |
visited = set() | |
q = deque([0]) | |
while q: | |
curr_node = q.pop() | |
visited.add(curr_node) | |
for conn_node in adj_list[curr_node]: | |
if conn_node not in visited: | |
q.append(conn_node) | |
return len(visited) | |
def criticalConnections(self, n: int, connections: List[List[int]]) -> List[List[int]]: | |
critical_conns = [] | |
# 1. remove one connection | |
for removed_conn in connections: | |
# 2. create adj list | |
adj_list = [[] for _ in range(n)] | |
for conn in connections: | |
if conn != removed_conn: | |
adj_list[conn[0]].append(conn[1]) | |
adj_list[conn[1]].append(conn[0]) | |
# 3. run bfs, calculate number of distinct visited nodes; | |
# if less than n, this is a critical connection | |
visited_number = self.bfsNodes(adj_list) | |
if visited_number < n: | |
critical_conns.append(removed_conn) | |
return critical_conns |
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