Lintcode 1088. 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.

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.

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Example

Example 1:

Input:  [[1,2], [1,3], [2,3]]
Output:  [2,3]	
Explanation: 
  looks like:
	  1
	 / \
	2 - 3

Example 2:

Input:  [[1,2], [2,3], [3,4], [1,4], [1,5]]
Output:  [1,4]	
Explanation:
	looks like:
	5 - 1 - 2
	    |   |
	    4 - 3

Code (Java):

 

public class Solution {
    /**
     * @param edges: List[List[int]]
     * @return: List[int]
     */
    public int[] findRedundantConnection(int[][] edges) {
        // write your code here
        int[] res = new int[2];
        
        if (edges == null || edges.length == 0) {
            return res;
        }
        
        UF uf = new UF(edges.length + 1);
        
        for (int[] edge : edges) {
            uf.union(edge[0], edge[1]);
        }
        
        return uf.getAns();
    }
}

class UF {
    private int[] parents;
    private int n;
    
    private int[] ans;
    
    public UF(int n) {
        this.n = n;
        parents = new int[n];
        ans = new int[2];
        
        for (int i = 0; i < n; i++) {
            parents[i] = i;
        }
    }
    
    public void union(int a, int b) {
        int pa = find(a);
        int pb = find(b);
        
        if (pa != pb) {
            parents[pa] = pb;
        } else {
            ans[0] = a;
            ans[1] = b;
        }
    }
    
    public int find(int a) {
        int root = a;
        while (parents[root] != root) {
            root = parents[root];
        }
        
        // path compression 
        while (root != a) {
            int parent = parents[a];
            parents[a] = root;
            a = parent;
        }
        
        return root;
    }
    
    public int[] getAns() {
        return ans;
    }
}