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684. Redundant Connection

Description

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.

Solutions

• Java
• C++
• Python
• Go
• Javascript

• 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];
      }
  }
  

• 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];
      }
  };
  

• class Solution:
      def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
          def find(x):
              if p[x] != x:
                  p[x] = find(p[x])
              return p[x]
  
          p = list(range(1010))
          for a, b in edges:
              if find(a) == find(b):
                  return [a, b]
              p[find(a)] = find(b)
          return []
  
  

• func findRedundantConnection(edges [][]int) []int {
  	p := make([]int, 1010)
  	for i := range p {
  		p[i] = i
  	}
  	var find func(x int) int
  	find = func(x int) int {
  		if p[x] != x {
  			p[x] = find(p[x])
  		}
  		return p[x]
  	}
  	for _, e := range edges {
  		a, b := e[0], e[1]
  		if find(a) == find(b) {
  			return e
  		}
  		p[find(a)] = find(b)
  	}
  	return []int{}
  }
  

• /**
   * @param {number[][]} edges
   * @return {number[]}
   */
  var findRedundantConnection = function (edges) {
      let p = Array.from({ length: 1010 }, (_, i) => i);
      function find(x) {
          if (p[x] != x) {
              p[x] = find(p[x]);
          }
          return p[x];
      }
      for (let [a, b] of edges) {
          if (find(a) == find(b)) {
              return [a, b];
          }
          p[find(a)] = find(b);
      }
      return [];
  };
  
  

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