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@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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