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66 lines
1.6 KiB
Plaintext
66 lines
1.6 KiB
Plaintext
---
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title: '684. Redundant Connection'
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description: In this problem, a tree is an undirected graph that is connected and has no cycles
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icon: dot
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topics:
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- { name: "Depth-First Search", slug: "depth-first-search" }
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- { name: "Breadth-First Search", slug: "breadth-first-search" }
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- { name: "Union-Find", slug: "union-find" }
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- { name: "Graph Theory", slug: "graph" }
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issue: "https://github.com/prdlk/leetcode/issues/146"
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sidebar:
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label: 'Redundant Connection'
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badge: 'Medium'
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---
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### Example 1:
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- Input: `edges = [[1,2],[1,3],[2,3]]`
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- Output: `[2,3]`
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### Example 2:
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- Input: `edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]`
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- Output: `[1,4]`
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### Constraints:
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- `n == edges.length`
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- `3 <= n <= 1000`
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- `edges[i].length == 2`
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- `1 <= ai < bi <= edges.length`
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- `ai != bi`
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- There are no repeated edges.
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- The given graph is connected.
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## Solution
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```py
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class Solution:
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def findRedundantConnection(self, edges: list[list[int]]) -> list[int]:
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N = len(edges)
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par = [i for i in range(N + 1)]
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rank = [1] * (N + 1)
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def find(n):
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if n != par[n]:
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par[n] = find(par[n])
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return par[n]
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def union(n1, n2):
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p1, p2 = find(n1), find(n2)
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if p1 == p2:
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return False
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if rank[p1] > rank[p2]:
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par[p2] = p1
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rank[p1] += rank[p2]
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else:
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par[p1] = p2
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rank[p2] += rank[p1]
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return True
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for n1, n2 in edges:
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if not union(n1, n2):
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return [n1, n2]
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```
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