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:
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.

Update (2017-09-26):
We have overhauled the problem description + test cases and specified clearly the graph is an undirected graph. For the directed graph follow up please see Redundant Connection II). We apologize for any inconvenience caused.

Problem URL

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Solution

给一个图，有N个节点，labeled from 1 to N。本来是一棵树，但是现在多了一条边，找到多的这条边。

We could maintain a parent array to denote the parent of a connected component. An edge will connect two nodes into one connected component.

When we count an edge in, if the two nodes have already been in the same connected component, the edge will result in a cycle. That is, the edge is redundant.

We can make use of Disjoint Sets (Union Find).
If we regard a node as an element, a connected component is actually a disjoint set.

Code

class Solution {
    public int[] findRedundantConnection(int[][] edges) {
        int[] parents = new int[edges.length + 1];
        for (int[] edge: edges){
            if (find(parents, edge[0]) == find(parents, edge[1])){
                return edge;
            }
            else{
                parents[find(parents, edge[0])] = find(parents, edge[1]);
            }
        }
        return new int[2];
    }
    
    private int find (int[] parents, int n){
        if (parents[n] == 0){
            return n;
        }
        else{
            return find(parents, parents[n]);
        }
    }
}

Time Complexity: O()
Space Complexity: O()

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