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

The given input is a graph that started as a tree with N nodes (with distinct values 1, 2, ..., 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 resulting graph is given as a 2D-array of edges. Each element of edges is a pair [u, v] with u < v, that represents an undirected edge connecting nodes u and v.

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 given 2D-array. The answer edge [u, v] should be in the same format, with u < v.

Example 1:

Input: [[1,2], [1,3], [2,3]]
Output: [2,3]
Explanation: The given undirected graph will be like this:
  1
 / \
2 - 3

Example 2:

Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]
Output: [1,4]
Explanation: The given undirected graph will be like this:
5 - 1 - 2
    |   |
    4 - 3

Note:

  • The size of the input 2D-array will be between 3 and 1000.
  • Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.

  • Update (2017-09-26):
    We have overhauled the problem description + test cases and specified clearly the graph is an undirected graph. For the directed graph follow up please see Redundant Connection II). We apologize for any inconvenience caused.

    Solution 1.

    // OJ: https://leetcode.com/problems/redundant-connection/
    // Author: github.com/lzl124631x
    // Time: O(N)
    // Space: O(N)
    class UnionFind {
    private:
      vector<int> id, rank;
      int cnt;
    public:
      UnionFind(int cnt) : cnt(cnt) {
        id = vector<int>(cnt);
        rank = vector<int>(cnt, 0);
        for (int i = 0; i < cnt; ++i) id[i] = i;
      }
      int find(int p) {
          if (id[p] == p) return p;
          return id[p] = find(id[p]);
      }
      int getCount() { return cnt; }
      bool connected(int p, int q) { return find(p) == find(q); }
      void connect(int p, int q) {
        int i = find(p), j = find(q);
        if (i == j) return;
        if (rank[i] < rank[j]) id[i] = j;
        else {
          id[j] = i;
          if (rank[i] == rank[j]) rank[j]++;
        }
        --cnt;
      }
    };
    
    class Solution {
    public:
        vector<int> findRedundantConnection(vector<vector<int>>& edges) {
            UnionFind uf(edges.size());
            for (auto e : edges) {
                if (uf.connected(e[0] - 1, e[1] - 1)) return e;
                uf.connect(e[0] - 1, e[1] - 1);
            }
        }
    };