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

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(*) 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 xs 0 = ([], xs)
split (x:xs) n = let (f,l) = split xs (n-1) in (x : f, l)
```

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))
```