Description
Count connected components in an undirected graph.
Constraints
- Time Complexity:
O(V+E) - Space Complexity:
O(V+E)
Tags
dfsbfsunion-findgraph
def countComponents(n, edges):
par = [i for i in range(n)]
rank = [1] * n
def find(n1):
res = n1
while res != par[res]:
par[res] = par[par[res]]
res = par[res]
return res
def union(n1, n2):
p1, p2 = find(n1), find(n2)
if p1 == p2:
return 0
if rank[p2] > rank[p1]:
par[p1] = p2
rank[p2] += rank[p1]
else:
par[p2] = p1
rank[p1] += rank[p2]
return 1
res = n
for n1, n2 in edges:
res -= union(n1, n2)
return res