Files
leetcode/apps/docs/content/(04-relational)/union-find/323-number-of-connected-components-in-an-undirected-graph.mdx
T

60 lines
1.4 KiB
Plaintext

---
title: '323. Number of Connected Components in an Undirected Graph'
description: You have a graph of n nodes. You are given an integer n and an array edges where edges[i] = [ai, bi] indicates that there is an edge between ai and bi in the graph
icon: dot
sidebar:
label: 'Number of Connected Components in an Undirected Graph'
badge: 'Medium'
---
<Badge variant="accent">Depth-First Search</Badge>
### Example 1:
- Input: `n = 5, edges = [[0,1],[1,2],[3,4]]`
- Output: `2`
### Example 2:
- Input: `n = 5, edges = [[0,1],[1,2],[2,3],[3,4]]`
- Output: `1`
### Constraints:
- `1 <= n <= 2000`
- `1 <= edges.length <= 5000`
- `edges[i]` = [ai, bi]
- `ai != bi`
- There are no repeated edges.
## Solution
```py
class Solution:
def countComponents(self, n: int, edges: list[list[int]]) -> int:
par = [i for i in range(n + 1)]
rank = [1] * (n + 1)
def find(n):
res = n
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[p1] > rank[p2]:
par[p2] = p1
rank[p1] += rank[p2]
else:
par[p1] = p2
rank[p2] += rank[p1]
return 1
res = n
for n1, n2 in edges:
res -= union(n1, n2)
return res
```