"""
684. Redundant Connection
Difficulty: Medium
https://leetcode.com/problems/redundant-connection/

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In this problem, a tree is an undirected graph that is connected and
has no cycles.

You are given a graph that started as a tree with n nodes labeled
from 1 to n, with one additional edge added. The added edge has two
different vertices chosen from 1 to n, and was not an edge that
already existed. The graph is represented as an array edges of length
n where edges[i] = [ai, bi] indicates that there is an edge between
nodes ai and bi in the graph.

Return an edge that can be removed so that the resulting graph is a
tree of n nodes. If there are multiple answers, return the answer that
occurs last in the input.



Example 1:

Input: edges = [[1,2],[1,3],[2,3]]
Output: [2,3]

Example 2:

Input: edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]
Output: [1,4]



Constraints:

        • n == edges.length

        • 3 <= n <= 1000

        • edges[i].length == 2

        • 1 <= ai < bi <= edges.length

        • ai != bi

        • There are no repeated edges.

        • The given graph is connected.
"""


class Solution:
    def findRedundantConnection(self, edges: list[list[int]]) -> list[int]:
        N = len(edges)
        par = [i for i in range(N + 1)]
        rank = [1] * (N + 1)

        def find(n):
            if n != par[n]:
                par[n] = find(par[n])
            return par[n]

        def union(n1, n2):
            p1, p2 = find(n1), find(n2)
            if p1 == p2:
                return False

            if rank[p1] > rank[p2]:
                par[p2] = p1
                rank[p1] += rank[p2]
            else:
                par[p1] = p2
                rank[p2] += rank[p1]

            return True

        for n1, n2 in edges:
            if not union(n1, n2):
                return [n1, n2]