Adding substance to the above critisizm, in your 'for loop' notice the following:

for i in range(0, 4):

will only check your remaining if, elif, else statements for the index 0 to 3 in arr1, arr2.

for i in range(0, len(arr1)):

will check those statements for the index 0 up to len(arr1) - 1 (this allows the program to scale with a possible change in list size)

We are told the list length will be the same so modifying your program with this new range in your for loop will suffice.

:P

you should see what sagemath can do. I strongly recommend using sagemath as it is python 3 with a huge library of extra modules specifically for applied and abstract mathematics.

lol FFS :P

This avoids having to use (factorial(n - 1) + 1) % (n^2) and bypasses huge divisions other than (factorial(n - 1) + 1)//(n^2) (which can be done before time out on most attempts), everything else is multiplication :P

Still checking if (factorial(n - 1) + 1) % (n^2) == 0 is the best option. Alternately, this shows modulo is not needed with some intuition about the floor of an integer and simple algebra.

lol clean the dojo! i am here doing kata's all night chilling and this post almost made me spit my drink out lolololol

[start:stop(not but not including): step] so str(string)[::-1] gives every character of the string by step -1, thus backwards.

LOL did the same thing :P dat python 3

Exactly what I did :P, the trick was making the last key 0 in the dictionary for the modulo to work for any nb_petals being a multiple of 6 :P

number % a == 0 and number % b == 0 evaluates to True if both are True
number % a == 0 and number % b == 0 evaluates to False if either or both are False

So there is actually a boolean being returned :P