684. Redundant Connection

• Difficulty: Medium

• Topics: Tree, Union Find, Graph

• Similar Questions:

  • Redundant Connection II

  • Accounts Merge

Problem:

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:

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.

Solutions:

class Solution {
public:
    class UF {
        public:
            bool isConnected(int a, int b) {
                return findParents(a) == findParents(b);
            }

            void connect(int a, int b) {
                int rootA = findParents(a);
                int rootB = findParents(b);
                parents[rootA] = rootB;
            }

            int findParents(int a) {
                if (parents.count(a) == 0 || parents[a] == a) { // check whether parents[a] == a; otherwise infinite resursion. 
                    parents[a] = a;
                } else {
                    parents[a] = findParents(parents[a]);
                }
                return parents[a];
            }

        private:
            unordered_map<int, int> parents;
    };

    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        UF uf;
        for (auto& edge : edges) {
            if (uf.isConnected(edge[0], edge[1]))   return edge;
            else {
                uf.connect(edge[0], edge[1]);
            }
        }

        return {};
    }
};