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## Learning Opportunities

This puzzle can be solved using the following concepts. Practice using these concepts and improve your skills.

## Goal

You are a beetle scientist and you have to identify the species of a lot of male beetles in your research.
Currently, you can easily identify these beetles after they are fully grown up.
However, in your research, you want to identify the species of the male beetles while they are still pupae.

Since beetles' pupae actually show the rough shape of the grown beetles,
you are able to measure the shape of their horn while they are still pupae.
Based on your observation, you found that there is a difference in the horn size of the pupae of different beetle species.
So you want to create a decision tree to identify the species of the beetles purely based on pupae horn sizes.

The tree is created by dividing one group into 2 groups repeatedly.
We split the group by a separator, which will be a pupa's horn length.
The length is selected to divide the group into 2 groups.

Starting from 1 single group with all beetle pupae,
step 1. we choice 1 beetle pupa in the group if there are multiple beetle species in the group
step 2. pupae with smaller horn sizes will be separated into a sub-group => step 1
step 3. all other pupae in the group will be separated into another sub-group => step 1

We scan through all pupae in the group and choice one beetle as a separator.
The equation of how we choose the pupa to be the separator is Binary Entropy.
And we find the best separator in the group with the smallest Weighted Entropy.

Entropy equations as this picture: https://imgur.com/gBWJ6fB.png

We keep creating branches for the binary tree until we have 0 entropy. (all pupae in a group is in 1 single species)

If 2 separators have the same entropy, we will always choose the pupa with a smaller index.
After we separate a group, we breakdown the group with smaller horns until we reach 0 entropy, and then we start working on the other group.

The output will be the index of the last separator (pupa) in the binary tree.

----
Given this input:

4
0 1 1
1 2 1
2 3 2
3 4 2

It shows that we have 4 pupae with their corresponding horn lengths from 1 to 4.
Since their index is from 0 to 3, we will name them pupa-0, pupa-1, pupa-2, and pupa-3.
--
If we select pupa-0 as the separator, nothing will change since there is no pupa have a smaller horn length then pupa-0.

If we select pupa-1 as the separator, pupa-0 will be separated into one sub-group, pupa-1,2,3 will be in another sub-group.
Pupa-0's group has a binary entropy of 0.
Pupa-1,2,3's group have a binary entropy of 0.9183.
Weighted entropy will be 0.6887.

If we select pupa-2 as the separator, pupa-0,1 will be separated into one sub-group, while pupa-2,3 will be in another sub-group.
Pupa-0,1's group has a binary entropy of 0, and pupa-2,3 as well.
Weighted entropy will be 0.

If we select pupa-3 as the separator, pupa-0,1,2 will be in one sub-group, and pupa-3 will be in another sub-group.
The resulting entropy will be 0.6887.
--
Select pupa-2 will have 0 entropy.
The output should be 2.
Input
Line 1: An integer N for the number of beetle pupae sorted by their horn sizes.
Next N lines: 3 space separated integers index, hornSize and speciesID for a point.
Output
Line 1: Last separator (pupa) index
Constraints
2 <= N <= 20
0 <= index < 20
0 <= hornSize <= 100
1 <= speciesID <= 4
Example
Input
```2
0 0 1
1 2 2```
Output
`1`

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