684. Redundant Connection

Description

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:

graph

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:

graph

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 directedgraph follow up please see Redundant Connection II). We apologize for any inconvenience caused.

Solution

Union-find

Disjoint Set (Or Union-Find) | Set 1 (Detect Cycle in an Undirected Graph)

class Solution {
    public int[] findRedundantConnection(int[][] edges) {
        int[] parent = new int[edges.length + 1];  // values are [1, n]! so size is n + 1
        Arrays.fill(parent, -1);
        
        for (int[] edge : edges) {
            int xset = find(parent, edge[0]);
            int yset = find(parent, edge[1]);
            if (xset == yset) {     // cicle detected
                return edge;
            }
            // union
            parent[xset] = yset;
        }
        
        return new int[2];
    }
    
    public int find(int[] parent, int i) {
        if (parent[i] == -1) {  // root
            return i;
        }
        parent[i] = find(parent, parent[i]);
        return parent[i];
    }
}