# 99 questions/Solutions/3

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(*) Find the K'th element of a list. The first element in the list is number 1.

This is (almost) the infix operator !! in Prelude, which is defined as:

```(!!)                :: [a] -> Int -> a
(x:_)  !! 0         =  x
(_:xs) !! n         =  xs !! (n-1)
```

Except this doesn't quite work, because !! is zero-indexed, and element-at should be one-indexed. So:

```elementAt :: [a] -> Int -> a
elementAt list i    = list !! (i-1)
```

Or without using the infix operator:

```elementAt' :: [a] -> Int -> a
elementAt' (x:_) 1  = x
elementAt' [] _     = error "Index out of bounds"
elementAt' (_:xs) k
| k < 1           = error "Index out of bounds"
| otherwise       = elementAt' xs (k - 1)
```

Alternative version:

```elementAt'' :: [a] -> Int -> a
elementAt'' (x:_) 1  = x
elementAt'' (_:xs) i = elementAt'' xs (i - 1)
elementAt'' _ _      = error "Index out of bounds"
```

This does not work correctly on invalid indexes and infinite lists, e.g.:

```elementAt'' [1..] 0
```

A few more solutions using prelude functions:

```elementAt'' xs n
| length xs < n = error "Index out of bounds"
| otherwise = fst . last \$ zip xs [1..n]

elementAt''' xs n = head \$ foldr (\$) xs
\$ replicate (n - 1) tail
-- Negative indices not handled correctly:
-- 'h'

elementAt'''' xs n
| length xs < n = error "Index out of bounds"
| otherwise = last \$ take n xs

elementAt''''' xs n
| length xs < n = error "Index out of bounds"
| otherwise = head . reverse \$ take n xs

elementAt'''''' xs n
| length xs < n = error "Index out of bounds"
| otherwise = head \$ drop (n - 1) xs
```

or `elementAt_w'` correctly in point-free style:

```elementAt_w'pf = (last .) . take . (+ 1)
```

Pedantic note: the above definition of `elementAt_w'pf` does not conform to the order of arguments specified by the question, but the following does:

```elementAt_w'pf' = flip \$ (last .) . take . (+ 1)
```