Redundant Connection

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.

class Solution {
    public int[] findRedundantConnection(int[][] edges) {
        HashMap<Integer, Integer> parent = new HashMap<>();
        for (int i = 1; i <= edges.length; i++)
            parent.put(i, i);

        for (int[] edge : edges) {
            int from = edge[0], to = edge[1];
            int fromParent = findTopParent(parent, from);
            int toParent = findTopParent(parent, to);
            if (fromParent == toParent)
                return edge;
            parent.put(fromParent, toParent);
        }

        return null;
    }

    private int findTopParent(HashMap<Integer, Integer> parent, int node) {
        while (node != parent.get(node))
            node = parent.get(node);
        return node;
    }
}