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82 lines
2.0 KiB
Python
82 lines
2.0 KiB
Python
"""
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684. Redundant Connection
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Difficulty: Medium
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https://leetcode.com/problems/redundant-connection/
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──────────────────────────────────────────────────
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In this problem, a tree is an undirected graph that is connected and
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has no cycles.
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You are given a graph that started as a tree with n nodes labeled
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from 1 to n, with one additional edge added. The added edge has two
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different vertices chosen from 1 to n, and was not an edge that
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already existed. The graph is represented as an array edges of length
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n where edges[i] = [ai, bi] indicates that there is an edge between
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nodes ai and bi in the graph.
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Return an edge that can be removed so that the resulting graph is a
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tree of n nodes. If there are multiple answers, return the answer that
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occurs last in the input.
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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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"""
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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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