[Haskell-beginners] Code review request?

[Ut Primum]

Hi,
this is how I would have solved the exercise:

canBalanceRec [] _ _ = False
canBalanceRec (x:xs) s1 s2 = s1+x==s2-x || canBalanceRec xs (s1+x) (s2-x)
canBalance xs = s==0 || canBalanceRec xs 0 s where s=sum xs

Note that there is one difference: in this case
canBalance [-2,-2,4] = True
because I think I can split the list into [ ] and the rest (since the sum
of the empty list is 0 and the same is the sum of the elements of the
lists). Anyway this is a matter of how we interpret the text of the
exercise (your function would return False instead). Note that removing the
"s==0 ||" makes no difference.

 An advantage of my solution is that it is less expensive, since its
computational complexity is linear in the length of the list xs. Yours uses
sum and drop many times, and so is slower. Just to have an idea of the
difference (using ghci interpreter):

> canBalance [1..10000]
False
(*0.02* secs, 6,974,552 bytes)
> can_split_even [1..10000]
False
(*5.60* secs, 11,395,843,384 bytes)

Cheers,
Ut

Il giorno mer 15 gen 2020 alle ore 18:27 Jake Vossen <jake at vossen.dev> ha
scritto:

> Hey everyone,
>
> Let me know if this is not the right place for this, but I am curious if
> someone could take a look at my code and maybe share some feedback / how
> a more experienced haskeller would approach this problem.
>
> New to Haskell, pretty experienced with imperative languages. I have
> solved the following problem in Haskell:
>
> > Given a non-empty array, return true if there is a place to split the
> > array so that the sum of the numbers on one side is equal to the sum
> > of the numbers on the other side.
>
> > canBalance([1, 1, 1, 2, 1]) -> true
> > canBalance([2, 1, 1, 2, 1]) -> false
> > canBalance([10, 10]) -> true
>
> Here is my code (my solution uses `can_split_even` not `canBalance`)
>
> ```
> can_split_even :: (Num a) => (Eq a) => [a] -> Bool
> can_split_even xs = True `elem` is_even_at_each_spot
>   where
>     is_even_at_each_spot :: [Bool]
>     is_even_at_each_spot = map (is_split xs) [1 .. (length xs - 1)]
>       where
>         is_split :: (Num a) => (Eq a) => [a] -> Int -> Bool
>         is_split xs index = sum (take index xs) == sum (drop index xs)
> ```
>
> Thanks so much!
>
> --
> Jake Vossen
> Colorado School of Mines, Class of 2022
> B.S. Computer Science
> PGP: 08CD 67DA FE3A 0AE7 B946  6AC3 B812 7052 D9E3 817B
> https://jake.vossen.dev
>
>
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