Algorithm to Remove a Redundant Connection from a Undirected Graph to Make a Valid Tree using Union-Find (Disjoint Set)

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.

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.

Remove a Redundant Connection using Disjoint Set (Union Find Algorithm

We can go through each edges and connect (join) the vertices using Disjoint Set Data Structure. If both vertices of an edge already belong to a group, which means there is a cycle, we can mark this edge redundant connection.

1
2
3
4
5
6
7
8
9
10
11
12

class Solution {
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        DSU dsu(edges.size() + 1);
        for (const auto &e: edges) {
            if (!dsu.join(e[0], e[1])) {
                return e;
            }
        }
        return {};
    }
};

class Solution {
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        DSU dsu(edges.size() + 1);
        for (const auto &e: edges) {
            if (!dsu.join(e[0], e[1])) {
                return e;
            }
        }
        return {};
    }
};

The Disjoint Set Data Structure is described and implemented here. With the path compression, the complexity for both space and time of this algorithm is O(N) where N is the number of the vertices in the undirected graph.

You might also solve this problem with O(N^2) using Depth First Search Algorithm.

–EOF (The Ultimate Computing & Technology Blog) —