In this problem, a tree is anundirectedgraph 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 ofedges. Each element ofedgesis a pair[u, v]withu < v, that represents anundirectededge connecting nodesuandv.

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, withu < 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 anundirectedgraph. For thedirectedgraph follow up please seeRedundant Connection II). We apologize for any inconvenience caused.

class Solution {
    private class UnionFind{
        int[] parent;

        public UnionFind(int n){
            parent = new int[n];

            for (int i = 0; i < n; i++){
                parent[i] = i;
            }
        }

        public int find(int x){
            if (x == parent[x]){
                return x;
            }

            return parent[x] = find(parent[x]);
        }

        public void union(int a, int b){
            int parent_a = find(a);
            int parent_b = find(b);

            if (parent_a != parent_b){
                parent[parent_a] = parent_b;
            }
        }
    }

    public int[] findRedundantConnection(int[][] edges) {
        if (edges == null || edges.length == 0) return new int[0];

        int n = edges.length;
        UnionFind uf = new UnionFind(n);

        for (int[] edge : edges){
            int parent_a = uf.find(edge[0] - 1);
            int parent_b = uf.find(edge[1] - 1);

            if (parent_a != parent_b){
                uf.union(edge[0] - 1, edge[1] - 1);
            }
            else{
                return edge;
            }
        }

        return new int[0];
    }
}