Task 1: Binary Flip

You are given a positive integer, $n.

Write a script to find the binary flip.
Example 1

Input: $n = 5
Output: 2

First find the binary equivalent of the given integer, 101.
Then flip the binary digits 0 -> 1 and 1 -> 0 and we get 010.
So Binary 010 => Decimal 2.

Example 2

Input: $n = 4
Output: 3

Decimal 4 = Binary 100
Flip 0 -> 1 and 1 -> 0, we get 011.
Binary 011 = Decimal 3

Example 3

Input: $n = 6
Output: 1

Decimal 6 = Binary 110
Flip 0 -> 1 and 1 -> 0, we get 001.
Binary 001 = Decimal 1


MY NOTES: very easy.  That's ones complement with a bit of range-fixing!

GUEST LANGUAGE: As a bonus, I also had a go at translating ch-1.pl
into C (look in the C directory for the translation)


Task 2: Equal Distribution

You are given a list of integers greater than or equal to zero, @list.

Write a script to distribute the number so that each members are same. If you succeed then print the total moves otherwise print -1.

Please follow the rules (as suggested by Neils van Dijke [2022-11-21 13:00]

1) You can only move a value of '1' per move
2) You are only allowed to move a value of '1' to a direct neighbor/adjacent cell


Example 1:

Input: @list = (1, 0, 5)
Output: 4

Move #1: 1, 1, 4
(2nd cell gets 1 from the 3rd cell)

Move #2: 1, 2, 3
(2nd cell gets 1 from the 3rd cell)

Move #3: 2, 1, 3
(1st cell get 1 from the 2nd cell)

Move #4: 2, 2, 2
(2nd cell gets 1 from the 3rd cell)

Example 2:

Input: @list = (0, 2, 0)
Output: -1

It is not possible to make each same.

Example 3:

Input: @list = (0, 3, 0)
Output: 2

Move #1: 1, 2, 0
(1st cell gets 1 from the 2nd cell)

Move #2: 1, 1, 1
(3rd cell gets 1 from the 2nd cell)


MY NOTES: Hmm..  If I'm understanding that right, it's two passes:
1. repeatedly find two adjacent cells whose absolute difference > 1
and transfer one from the bigger cell to the smaller cell.
2. repeatedly find 2 adjacent cells where firstvalue < secondvalue
and transfer one from secondvalue to firstvalue

Note also that I'm pretty sure that all lists whose sum is divisible
by 3 can be balanced, and no list whose sum is NOT divisible by 3 can be.

GUEST LANGUAGE: As a bonus, I also had a go at translating ch-2.pl
into C (look in the C directory for the translation)