684 Redundant Connection

684. Redundant Connection

1. Question

n 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 ofedges. Each element ofedgesis a pair[u, v]withu < v, that represents an undirected edge 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.

2. Implementation

(1) Union Find

class Solution {
    public int[] findRedundantConnection(int[][] edges) {
        int n = edges.length;
        UnionFind uf = new UnionFind(n);

        for (int[] edge : edges) {
            if (!uf.union(edge[0], edge[1])) {
                return new int[] {edge[0], edge[1]};
            }
        }
        return new int[0];
    }

    class UnionFind {
        int[] sets;
        int[] size;
        int count;

        public UnionFind(int n) {
            sets = new int[n + 1];
            size = new int[n + 1];
            count = n;

            for (int i = 0; i <= n; i++) {
                sets[i] = i;
                size[i] = 1;
            }
        }

        public int find(int node) {
            while (node != sets[node]) {
                node = sets[node];
            }
            return node;
        }

        public boolean union(int i, int j) {
            int node1 = find(i);
            int node2 = find(j);

            if (node1 == node2) {
                return false;
            }

            if (size[node1] < size[node2]) {
                sets[node1] = node2;
                size[node2] += size[node1];
            }
            else {
                sets[node2] = node1;
                size[node1] += size[node2];
            }

            --count;
            return true;
        }
    }
}

3. Time & Space Complexity

Union Find: 时间复杂度O(n), 空间复杂度O(n)