https://programmers.co.kr/learn/courses/30/lessons/49189
Code
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from collections import defaultdict
from queue import deque
# w - f - i
def bfs(graph, src):
ds = {src: 0}
q = deque([src])
while q:
cur_v = q.popleft()
for next_v in graph[cur_v]:
if next_v not in ds:
ds[next_v] = ds[cur_v] + 1
q.append(next_v)
return ds
def solution(n, vertex):
graph = defaultdict(set)
for v1, v2 in vertex:
graph[v1].add(v2)
graph[v2].add(v1)
ds = bfs(graph, 1)
max_val = max(ds.values())
return len([v for v, d in ds.items() if d == max_val])
Complexity
$O(|V||E|)$
- $V$ =
vertices
- $E$ =
edges
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