684. Redundant Connection.md

684. Redundant Connection

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leetcode-cn Daily Challenge on January 13th, 2021.

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Difficulty : Medium

Related Topics : Tree、Union Find、Graph

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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.

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Solution

• mine
  • Java
    • UnionFind Runtime: 1 ms, faster than 64.33%, Memory Usage: 39.6 MB, less than 36.58% of Java online submissions

      // O(N)time
      // O(N)space
      public int[] findRedundantConnection(int[][] edges) {
          int[] res = new int[2];
          int max = 0;
          for(int[] e : edges){
              max = Math.max(e[1], max);
          }
          UF uf = new UF(max);
          for(int[] e : edges){
              if(!uf.union(e[0] - 1, e[1] - 1)){
                  res = e;
              }
          }
          return res;
      }
      
      class UF{
          int[] arr;
          int count;
      
          public UF(int n){
              count = n;
              arr = new int[n];
              for(int i = 0; i < n; i++){
                  arr[i] = i; 
              }
          }
      
          boolean union(int a, int b){
              if(find(a) != find(b)){
                  arr[find(a)] = b;
                  count--;
                  return true;
              }
              return false;
          }
      
          int find(int a){
              if(arr[a] != a){
                  arr[a] = find(arr[a]);
              }
              return arr[a];
          }
      }
      

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• the most votes

• UnionFind Runtime: 1 ms, faster than 70.87%, Memory Usage: 40 MB, less than 49.64% of Java online submissions

  // O(N)time
  // O(N)space
  public int[] findRedundantConnection(int[][] edges) {
      int[] parent = new int[2001];
      for (int i = 0; i < parent.length; i++) parent[i] = i;
  
      for (int[] edge: edges){
          int f = edge[0], t = edge[1];
          if (find(parent, f) == find(parent, t)) return edge;
          else parent[find(parent, f)] = find(parent, t);
      }
  
      return new int[2];
  }
  
  private int find(int[] parent, int f) {
      if (f != parent[f]) {
        parent[f] = find(parent, parent[f]);
      }
      return parent[f];
  }
  

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