Hi there!

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Bitwise_AND

This explonation was took from 'JohanWiltink'

It evaluates as (a & b) & ((a & b) - 1).

Let c = a & b, so you have c & c-1. That clears the top bit, if any, from c. If c == 0, a and b had no set bits in common. Otherwise, if c & c-1 == 0, a and b had exactly one set bit in common. In all other cases, they had at least two set bits in common, and the function should return true.

You can use c & c-1 iteratively to zero out all set bits from c. That's a faster way to count them than to count set bits over all of them.

Hmmm... now I see... NO THING! HA!

a poor solution is to push multiples into an array, which fills the memory buffer and crashes. then, using a counter took me further, but that timed out after 12 seconds. finally, I tried a generator to economize memory usage. but the way a generator is structured, I was unable to pass n to solution() without hardcoding the n value.

You don't have to use 2 reduces regardless. You can access array elements with their index. You are doing a bunch of unnecessary stuff like concat and that ternary...

There are 263 javascript solutions ranging from 1 to 50+ lines. There are many valid ways to solve this one.

'Y' is sometimes considered a semivowel, but when only two categories (vowels / consonants) are considered, 'Y' is usually categorized as consonants.