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