Insert an element at a given position into a list.
insertAt :: a -> [a] -> Int -> [a] insertAt x xs (n+1) = let (ys,zs) = split xs n in ys++x:zs
insertAt :: a -> [a] -> Int -> [a] insertAt x ys 1 = x:ys insertAt x (y:ys) n = y:insertAt x ys (n-1)
As a note to the above solution - this presumes that the inserted argument will be a singleton type a inserted into a list [a]. The lisp example does not infer this intent. As a result, presuming the data to be inserted is likewise of type [a] (which we are tacitly inferring here to be String into String insertion), a solution is:
insertAt x xs n = take (n-1) xs ++ x ++ drop (n-1) xs
This solution, like many others in this quiz presumes counting element positions starts at 1, perhaps causing needless confusion.