In the simple QUICK UNION, we just join the components without any calculation. In the WEIGHTED QUICK UNION, we calculate the number of components in the two parts before joining.
We always place them in an intelligent way as explained in the below figure:
Hence it gives a better performance.
The solution to the given problem is as follows:

===

Hence. we can see that there are 3 trees in the forest after all the unions have been performed.
The id[] array contents are as follows:
|
component |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
|
id value |
3 |
3 |
3 |
3 |
3 |
3 |
3 |
7 |
7 |
9 |
9 |
9 |
9 |
===================================================================
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quick-union smaller tree smaller larger tree tree larger tree might put the larger tree lower
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Union(2,10) - So each key above is in its own set so if we
perform union(2,10) im assuming it should make a set that contains
both key such as s1 = {2, 10} or using an array for
example
-1
-1
2
10
-1 meaning vertices in own set then this should result in after
union as
-2
2
2
10
-2 being the number of vertices, positive 2 being the parent of
10...
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