Union find algorithm - Julia asked the peer to give a different question to ask - January 8, 2018 - 10:00 PM mock interview

UnionFind_SimilarAlgorithm discussion

/*
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

third edge -> 

Example 2:  distinct 5, 5 distinct edge, 4 distinct edge 
Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]
Output: [1,4]
Explanation: The given undirected graph will be like this:

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.

---- Discussion starts from here --- 
Julia's idea and the discussion with the peer:
Determine if there is a cycle. How to determine if there is a cycle? 
Tree, graph, using depth first search. 


Union find algorithm - one group 

The peer asked Julia what is the base case. 

What is the base case?

  1
 / \
2 - 3

3 -> 2 -> 1

base case: 
    
    4
   /  \
   1   3
   /\
  5  2
   
   
Hint given by the peer: 

Example 1: Hint given by peer: 
[1,2]
set(1,2)

[1,3]
set(1,2,3)

[2,3]

Example 2:
set(1), set(2), set(3), set(4), set(5)

1. [1,2]
set(1, 2)

2. [2,3]
set(1, 2, 3)

3. [3,4]
set(1,2,3,4)

4. [1,4]

It is called union-find merge disjointed groups.

5 - 1 - 2
    |   |
    4 - 3
    
[1, 4] in part of cycle 
    
detect the cycle 
    
5 -> 1 -> 2 -> 3 -> 4 -> 1, there is cycle 
4 -> 1 -> 5

*/