The task is really easy but the description is incomprensible.

Here is what is to be done, in, I hope, clearer words: You're given two integers. Let m be the largest and n the smallest. While m is superior to 0, at each step you must divide m by 2 (integer division), if there is rest on the division, you must add the product of n by the rest to a list (since the rest is always 1, it is equivalent to simply add n to the list), then multiply n by 2 in all cases. Eventually return the reversed list.

It's a way in Python to reverse a value, so basically it's doing a boolean evaluation (True/False) against whether the string values are the same forward and backwards, a.k.a a "Palindrome".

The task is really easy but the description is incomprensible.

Here is what is to be done, in, I hope, clearer words: You're given two integers. Let m be the largest and n the smallest. While m is superior to 0, at each step you must divide m by 2 (integer division), if there is rest on the division, you must add the product of n by the rest to a list (since the rest is always 1, it is equivalent to simply add n to the list), then multiply n by 2 in all cases. Eventually return the reversed list.

I would rate it harder if this challenge could not be bypassed without regex

Please use new python test framework.

It's not necessary to check (I+j)% == 0

I don't understand the question.

oh ok, I guess there is another method then! I can't think of any though hmm, anyway thanks!

Why loop though?

C: there is timeout error with a very simple loop (it passes the tests). Can anyone confirm if it is solvable?

It's a way in Python to reverse a value, so basically it's doing a boolean evaluation (True/False) against whether the string values are the same forward and backwards, a.k.a a "Palindrome".

https://docs.python.org/3/library/functions.html#slice

Can you please explain what is [::-1] doing ?

count the number of 1s in the array. Let this count be n. The Stanton measure is the number of times that n appears in the array.

Just try to understand this :)

in Simple english - whatever the occurnce of 1 come in array will be x and then find that occurnce of x in array

I got it, what it's mean--- never using array at all. Using operation inside memory

I got the exact same, @lepasi (:

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