oh i see. Thanks for the info. I will read about it.

This version and an equivalent full function do the exact same thing; this is just compressing it into one line. There's literally no advantage and it violates PEP-8 but some people seem to think it looks cool.

a nice tap!

This is cool I didn't know you could do that.

genius

wtf so there is a pattern?!

According to the description, where is my money? 😉

Yes, you can chain as many as you like

wait what, we can make 2 if statement in one line???

this kata makes me feel bad :(

men i love using lambda ;)

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

Can someone please explain what is happening here?
And why the time complexity of this solution is O(n)?