Question

684. Redundant Connection

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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

Solution 1

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 []

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];
  }
};

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 [];
};