684. Redundant Connection.md

684. Redundant Connection

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