684. Redundant Connection

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.

class Solution {
    unordered_map<int, int> group;
    int getGroup(int n) {
        return group.count(n) ? _getGroup(n) : group[n] = n;
    }

    int _getGroup(int n) {
        return n == group[n] ? n : group[n] = _getGroup(group[n]);
    }
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        for(auto edge : edges) {
            if(getGroup(edge.front()) ^ getGroup(edge.back())) {
                group[getGroup(edge.front())] = group[getGroup(edge.back())] = min(group[getGroup(edge.front())], group[getGroup(edge.back())]);
            } else {
                return edge;
            }
        }
        return {-1, -1};
    }
};