99 questions/Solutions/21

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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)

There are two possible simple solutions. First we can use split from problem 17 (or even splitAt from the Prelude) to split the list and insert the element. Second we can define a recursive solution on our own.

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.

A solution using foldl and a closure, also assumes lists are 1 indexed:

insertAt :: a -> [a] -> Int -> [a]
insertAt el lst n = fst $ foldl helper ([],1) lst
    where helper (acc,i) x = if i == n then (acc++[el,x],i+1) else (acc++[x],i+1)

The use of foldl imposes the use of concatenation. With a foldr we can use (:) instead, which is faster (O(n) vs. O(n²)). The use of zip [1..] does not seem to add any overhead compared to the same solution with the index stored in the accumulator.

insertAt :: a -> [a] -> Int -> [a]
insertAt elt lst pos = foldr concat' [] $ zip [1..] lst
        concat' (i, x) xs
            | i == pos  = elt:x:xs
            | otherwise = x:xs

Compared to the simple recursive definition, the fold version visits every elements of the list, whereas we could just stop after insertion of the element.