684. Redundant Connection

In this problem, a tree is an undirected graph that is connected and has no cycles.

You are given a graph that started as a tree with n nodes labeled from 1 to 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 graph is represented as an array edges of length n where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the graph.

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 input.

 Example 1:

Input: edges = [[1,2],[1,3],[2,3]]
Output: [2,3]

Example 2:

Input: edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]
Output: [1,4]

 Constraints:

• n == edges.length
• 3 <= n <= 1000
• edges[i].length == 2
• 1 <= ai < bi <= edges.length
• ai != bi
• There are no repeated edges.
• The given graph is connected.

JAVA

• class Solution {
      private int[] p;
  
      public int[] findRedundantConnection(int[][] edges) {
          p = new int[1010];
          for (int i = 0; i < p.length; ++i) {
              p[i] = i;
          }
          for (int[] e : edges) {
              int a = e[0], b = e[1];
              if (find(a) == find(b)) {
                  return e;
              }
              p[find(a)] = find(b);
          }
          return null;
      }
  
      private int find(int x) {
          if (p[x] != x) {
              p[x] = find(p[x]);
          }
          return p[x];
      }
  }

C++

• class Solution {
  public:
      vector<int> p;
  
      vector<int> findRedundantConnection(vector<vector<int>>& edges) {
          p.resize(1010);
          for (int i = 0; i < p.size(); ++i) p[i] = i;
          for (auto& e : edges) {
              int a = e[0], b = e[1];
              if (find(a) == find(b)) return e;
              p[find(a)] = find(b);
          }
          return {};
      }
  
      int find(int x) {
          if (p[x] != x) p[x] = find(p[x]);
          return p[x];
      }
  };

PYTHON

• class Solution:
      def findRedundantConnection(self, edges):
          p = list(range(1010))
  
          def find(x):
              if p[x] != x:
                  p[x] = find(p[x])  # Path compression
              return p[x]
  
          for a, b in edges:
              if find(a) == find(b):
                  return [a, b]
              p[find(a)] = find(b)
  
          return None