how do you even know something like that.....

I think the count(n)=count(s) solution is cleaner and slightly more efficient. It makes 41 comparisons whereas this one makes 43 (2 more in last line) and creates var x,y and sets them to 0.

So is this more efficient than using list.count 4 times?

This is wrong. The random trials do not append 0s to the end of the original array. See @Hendec 's comment

Yes, this is very broken and the instructions are unclear besides. Why 0.0 --> 0 but '0' --> '0' is beyond me.

I didn't get it at first either. The length of signature is getting larger and increasing by 1 each iteration. Once the length of signature grows to > n, the recursion terminates and returns all values in the now-large signature up to n-1. There's one extra calculation of the sum it's cool. Also handles the n<=3 cases nicely.

Can someone explain this?