# Difference between revisions of "99 questions/Solutions/21"

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− | insertAt x xs n = take (n-1) xs ++ x ++ drop (n-1) xs |
+ | insertAt x xs n = take (n-1) xs ++ [x] ++ drop (n-1) xs |

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</haskell> |
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## Revision as of 22:21, 4 January 2012

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
```

or

```
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)
```