when whould smash' be called?

O(n log n) time
O(1) space because no extra data structures used

this is best

O(n) time
O(1) space

O(n^2) time because String.sub creates a new string for each element of the input

Nice

O(List.length list) time complexity
O(List.length lr) space complexity due to stack space from reverse

O(String.length s) time complexity
O(1) space complexity bc aux is tail recursive

I think this is O(n^2) because String.sub creates a new string on each call.

I had the same idea. I avoid creating a new string by keeping track of two pointers, beginning and end.

what is ~?

what's the @ symbol do?

My google search didn't turn up anything apparent

EasterEgg should not return <li key={key}>{name}</li>?