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

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

This solution has O(String.length s) space complexity because String.lowercase_ascii creates a new string.

Thanks for the pointer (no pun intended)! In hindsight that's a clear optimisation.

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.