that vary interstent !!

Yes, you can chain as many as you like

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

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)?

I like to see everything in one line. But it's not good practice.

Nice observation skills and explanation. I Thank You for your time.

Still broken the same way. Could not complete.