(*) Split a list into two parts; the length of the first part is given.
Do not use any predefined predicates.Solution using
split xs n = (take n xs, drop n xs)
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)