a nice tap!

It does the trick but there are some really concise solutions here with a fraction of the upvotes...

learnes a big lesson

Nice way to evaluate the arr is not empty or None.

Sorted is O(n log n) complexity, both sorted and sort use the same algorithm behind the scenes. Creating a list is O(1).

The only difference with my solution is math library

coooooool

L comment I won't lie

It can also be None according to the kata description, which would then return None and fail.

intersting!!!

Old comment, but if someone runs into a similar issue - are you returning an array of the values or just the values?

My first attempt returned 10,-65 instead of [10,-65] for the first example problem, which fails.

i did same thing but i just said arr != None its acccepted too here >1

really nice

Since every character in the string has to be touched, assuming 'n' refers to the size of 'aeiouAEIOU' -> 10

String comparison method:

vowel, equal frequencies: 5.5 checks

not a vowel: 10 checks, constant

Set, assuming underlying b-tree has minimal depth:

vowel, equal frequency: 1 * .1 + 2 * .2 + 3 + .4 + 4 * .3 = 2.9 checks

not a vowel: upper bound of 4 checks, lower bound 3 checks

Comparing: vowel 5.5 vs 2.9 -> ~1.90 faster for set

not a vowel: 10 vs 3 to 4 -> 3.33 to 2.5 times faster, in theory

Note: this is in theory, the python interpreter and execution environment could make this worse or better

Obviously as 'n' gets (not much) larger, the set is going to be faster. For n = 10, I would say tradeoff is speed versus speed of coding