684. Redundant Connection

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.

Solution in python:

class Solution:
    def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
        def find(arr, x):
            if arr[x] == x:
                return x
            else:
                return find(arr, arr[x])
        uset = [i for i in range(1001)]
        for edge in edges:
            value1 = find(uset, edge[0])
            value2 = find(uset, edge[1])
            if value1 != value2:
                uset[value1] = value2
            else:
                return edge

Complexity analysis:

• Time complexity: O(E) <= T <= O(E^2)
• Space complexity: O(N)