# Difference between revisions of "99 questions/Solutions/17"

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where helper (i,(left,right)) x = if i >= n then (i+1,(left,right++[x])) else (i+1,(left++[x],right)) |
where helper (i,(left,right)) x = if i >= n then (i+1,(left,right++[x])) else (i+1,(left++[x],right)) |
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</haskell> |
</haskell> |
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+ | |||

+ | A solution that dequeues onto a stack and then reverses at the end: |
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+ | <haskell> |
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+ | split :: [a] -> Int -> ([a], [a]) |
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+ | split xs n = let (a, b) = helper [] xs n in (reverse a, b) |
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+ | where helper left right@(r:rs) n |
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+ | | n == 0 = (left, right) |
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+ | | otherwise = helper (r:left) rs (n - 1) |
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+ | |||

+ | </haskell> |
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+ | |||

+ | --- |
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+ | |||

+ | A recursive solution constructing the 2-tuple: |
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+ | |||

+ | <haskell> |
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+ | split :: [a] -> Int -> ([a],[a]) |
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+ | |||

+ | split [] _ = ([],[]) |
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+ | |||

+ | split (x:xs) n |
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+ | | n > 0 = ( |
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+ | x: (fst (split xs (n-1))), |
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+ | snd (split xs (n-1)) |
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+ | ) |
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+ | | n <= 0 = ( |
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+ | fst (split xs 0), |
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+ | x:(snd (split xs 0)) |
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+ | ) |
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+ | </haskell> |
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+ | |||

+ | A simple solution using recursion: |
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+ | <haskell> |
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+ | splitAt3 :: Int -> [a] -> ([a], [a]) |
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+ | splitAt3 n xs = if n < 0 then ([], xs) else splitR n xs [] |
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+ | where |
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+ | splitR 0 xs accum = (reverse accum, xs) |
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+ | splitR _ [] accum = (reverse accum, []) |
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+ | splitR n (x:xs) accum = splitR (n-1) xs (x : accum) |
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+ | </haskell> |
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+ | |||

+ | [[Category:Programming exercise spoilers]] |

## Latest revision as of 21:07, 10 March 2019

(*) Split a list into two parts; the length of the first part is given.

Do not use any predefined predicates.

Solution using `take`

and `drop`

:

```
split xs n = (take n xs, drop n xs)
```

Or even simpler using `splitAt`

:

```
split = flip splitAt
```

But these should clearly be considered "predefined predicates". Alternatively, we have the following recursive solution:

```
split :: [a] -> Int -> ([a], [a])
split [] _ = ([], [])
split l@(x : xs) n | n > 0 = (x : ys, zs)
| otherwise = ([], l)
where (ys,zs) = split xs (n - 1)
```

The same solution as above written more cleanly:

```
split :: [a] -> Int -> ([a], [a])
split (x:xs) n | n > 0 = let (f,l) = split xs (n-1) in (x : f, l)
split xs _ = ([], xs)
```

Or (ab)using the "&&&" arrow operator for tuples:

```
split :: [a] -> Int -> ([a], [a])
split (x:xs) n | n > 0 = (:) x . fst &&& snd $ split xs (n - 1)
split xs _ = ([], xs)
```

A similar solution using foldl:

```
split :: [a] -> Int -> ([a], [a])
split [] _ = ([], [])
split list n
| n < 0 = (list, [])
| otherwise = (first output, second output)
where output = foldl (\acc e -> if third acc > 0 then (first acc ++ [e], second acc, third acc - 1) else (first acc, second acc ++ [e], third acc)) ([], [], n) list
```

Note that for the above code to work you must define your own first, second, and third functions for tuples containing three elements like so:

```
first :: (a, b, c) -> a
first (x, _, _) = x
second :: (a, b, c) -> b
second (_, y, _) = y
third :: (a, b, c) -> c
third (_, _, z) = z
```

Another foldl solution without defining tuple extractors:

```
split :: [a] -> Int -> ([a],[a])
split lst n = snd $ foldl helper (0,([],[])) lst
where helper (i,(left,right)) x = if i >= n then (i+1,(left,right++[x])) else (i+1,(left++[x],right))
```

A solution that dequeues onto a stack and then reverses at the end:

```
split :: [a] -> Int -> ([a], [a])
split xs n = let (a, b) = helper [] xs n in (reverse a, b)
where helper left right@(r:rs) n
| n == 0 = (left, right)
| otherwise = helper (r:left) rs (n - 1)
```

---

A recursive solution constructing the 2-tuple:

```
split :: [a] -> Int -> ([a],[a])
split [] _ = ([],[])
split (x:xs) n
| n > 0 = (
x: (fst (split xs (n-1))),
snd (split xs (n-1))
)
| n <= 0 = (
fst (split xs 0),
x:(snd (split xs 0))
)
```

A simple solution using recursion:

```
splitAt3 :: Int -> [a] -> ([a], [a])
splitAt3 n xs = if n < 0 then ([], xs) else splitR n xs []
where
splitR 0 xs accum = (reverse accum, xs)
splitR _ [] accum = (reverse accum, [])
splitR n (x:xs) accum = splitR (n-1) xs (x : accum)
```