684. Redundant Connection

Problem:

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.

Analysis：

See graph valid tree.

Solution:

class Solution {
    public int[] findRedundantConnection(int[][] edges) {
        if (edges == null || edges.length == 0|| edges[0].length == 0)
            return null;
        
        int[] roots = new int[edges.length + 1];
        for (int i = 1; i <= edges.length; i++) {
            roots[i] = i;
        }
        for (int[] edge: edges) {
            int x = find(roots, edge[0]);
            int y = find(roots, edge[1]);
            if (x == y)
                return new int[]{edge[0], edge[1]};
            roots[x] = y;
        }
        return null;
    }

    private int find(int[] roots, int v) {
        if (roots[v] != v) {
            roots[v] = find(roots, roots[v]);
        }
        return roots[v];
    }
}