Question:

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.

Solution:

  • Method 1: dfs
    • how to find a cycle?
      • before we add current path (u, v) to the graph, we can already find the path from u to v, so after adding current edge, there exists a cycle. 
          class Solution {
  private Map<Integer, List<Integer>> graph;
  public int[] findRedundantConnection(int[][] edges) {
  graph = new HashMap<>();
  for(int[] edge : edges){
      Set<Integer> visited = new HashSet<>();
      if(hasPath(edge[0], edge[1], visited))
          return edge;
      List<Integer> temp = graph.containsKey(edge[0]) ? graph.get(edge[0]): new ArrayList<>();
      temp.add(edge[1]);
      graph.put(edge[0], temp);
      temp = graph.containsKey(edge[1]) ? graph.get(edge[1]): new ArrayList<>();
      temp.add(edge[0]);
      graph.put(edge[1], temp);
  }
  return new int[]{-1, -1};
  }
  private boolean hasPath(int u, int v, Set<Integer> visited){
  if(u == v) return true;
  visited.add(u);
  if(!graph.containsKey(u) || !graph.containsKey(v)) return false;
  List<Integer> nexts = graph.get(u);
  for(Integer next : nexts){
      if(visited.contains(next)) continue;
      if(hasPath(next, v, visited)) return true;
  }
  return false;
  }
  }
  • Method 2: Union find set 
    class Solution {
    private int[] uf;
    public int[] findRedundantConnection(int[][] edges) {
        uf = new int[edges.length * 2 + 1];
        for(int i = 1; i < uf.length; i++){
            uf[i] = i;
        }
        for(int[] edge : edges){
            int p = find(edge[0]);
            int q = find(edge[1]);
            if(p == q) return edge;
            uf[p] = q;
        }
        return null;
    }
    private int find(int i){
        if(i != uf[i]){
            uf[i] = find(uf[i]);
        }
        return uf[i];
    }
}

Reference

  1. 花花酱 LeetCode 684. Redundant Connection