684

684. Redundant Connection (Medium)

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.

Solution 1.

// OJ: https://leetcode.com/problems/redundant-connection/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class UnionFind {
private:
  vector<int> id, rank;
  int cnt;
public:
  UnionFind(int cnt) : cnt(cnt) {
    id = vector<int>(cnt);
    rank = vector<int>(cnt, 0);
    for (int i = 0; i < cnt; ++i) id[i] = i;
  }
  int find(int p) {
      if (id[p] == p) return p;
      return id[p] = find(id[p]);
  }
  int getCount() { return cnt; }
  bool connected(int p, int q) { return find(p) == find(q); }
  void connect(int p, int q) {
    int i = find(p), j = find(q);
    if (i == j) return;
    if (rank[i] < rank[j]) id[i] = j;
    else {
      id[j] = i;
      if (rank[i] == rank[j]) rank[j]++;
    }
    --cnt;
  }
};

class Solution {
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        UnionFind uf(edges.size());
        for (auto e : edges) {
            if (uf.connected(e[0] - 1, e[1] - 1)) return e;
            uf.connect(e[0] - 1, e[1] - 1);
        }
    }
};