First day here! gained plenty of knowledge today
Glad i stumbled across this site

.

right... xD

I went for counter object, so dont beat yourself up

Take a breath

def solution(s):

It started on the wrong foot. Yet, we all know what it is. We all know that C is a character in string. In real world, you're correct. This isn't the real world. This isn't 100 lines of code, with nested functions, loops, etc. But yes, you have a point.

Why is it the cleanest? Since when one and two letter variable names became the clean code solution?

_ = "abcdefghijklmnopqrstuvwxyz"
__   = _[7]   # h
___  = _[4]   # e
____ = _[11]  # l

print(__ + ___ + ____ * 2)  # hell

Then it's $ (n - 2)^3 $ and it depends on $ n $ value. Added that term to the formula.

I get to reuse my code from a current Beta kata:

n = 3 case

1st cube:

/\_\
\/_/

2nd cube:

 /\_\_\
/\/\_\_\
\/\/_/_/
 \/_/_/
 
3rd cube:

  /\_\_\_\
 /\/\_\_\_\
/\/\/\_\_\_\
\/\/\/_/_/_/
 \/\/_/_/_/
  \/_/_/_/

so then the finaly configuration is obviously where you stack cube #2 on top of #3. Then stack #1 on top of #2.

What do you call cube #3? if it's the third from the top, it's $ 3^3 = 27 $

By rethinking how to approach this problem.

Do you actually need to count from 1 to a million to know that there are indeed a million integers from one to a million? You could, sure, but I really hope you didn't answer yes.

Now think about the intervals [0, 10] and [5, 15]: they can be reduced to [0, 15], but why ?

Words in Binary with each letter from "hello world!" being a letter in each word