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

Still broken the same way. Could not complete.