# 99 questions/1 to 10

This is part of Ninety-Nine Haskell Problems, based on Ninety-Nine Prolog Problems and Ninety-Nine Lisp Problems.

If you want to work on one of these, put your name in the block so we know someone's working on it. Then, change n in your block to the appropriate problem number, and fill in the <Problem description>,<example in lisp>,<example in Haskell>,<solution in haskell> and <description of implementation> fields.

## Problem 1

(*) Find the last box of a list.

Example:

* (my-last '(a b c d)) (D)

Example in Haskell:

```
Prelude> myLast [1,2,3,4]
[4]
Prelude> myLast ['x','y','z']
"z"
```

Solution:

```
myLast :: [a] -> [a]
myLast [x] = [x]
myLast (_:xs) = myLast xs
```

Haskell also provides the function `last`

.

## Problem 2

(*) Find the last but one box of a list.

Example:

* (my-but-last '(a b c d)) (C D)

Example in Haskell:

```
Prelude> myButLast [1,2,3,4]
[3,4]
Prelude> myButLast ['a'..'z']
"yz"
```

Solution:

```
myButLast :: [a] -> [a]
myButLast list = drop ((length list) - 2) list
```

This simply drops all the but last two elements of a list.

Some other options:

```
myButLast = reverse . take 2 . reverse
```

or

```
myButLast = last . liftM2 (zipWith const) tails (drop 1)
```

Remark: The Lisp solution is actually wrong, it should not be the last two elements; a correct Haskell solution is:

```
myButLast = last . init
Prelude> myButLast ['a'..'z']
'y'
```

See also the solution to problem 2 in the Prolog list.

## Problem 3

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

Example:

* (element-at '(a b c d e) 3) C

Example in Haskell:

```
Prelude> elementAt [1,2,3] 2
2
Prelude> elementAt "haskell" 5
'e'
```

Solution:

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)
```

## Problem 4

(*) Find the number of elements of a list.

Example in Haskell:

```
Prelude> length [123, 456, 789]
3
Prelude> length "Hello, world!"
13
```

Solution:

```
length :: [a] -> Int
length [] = 0
length (_:l) = 1 + length l
```

This function is defined in Prelude.

## Problem 5

(*) Reverse a list.

Example in Haskell:

```
Prelude> reverse "A man, a plan, a canal, panama!"
"!amanap ,lanac a ,nalp a ,nam A"
Prelude> reverse [1,2,3,4]
[4,3,2,1]
```

Solution: (defined in Prelude)

```
reverse :: [a] -> [a]
reverse = foldl (flip (:)) []
```

The standard definition is concise, but not very readable. Another way to define reverse is:

```
reverse :: [a] -> [a]
reverse [] = []
reverse (x:xs) = reverse xs ++ [x]
```

## Problem 6

(*) Find out whether a list is a palindrome. A palindrome can be read forward or backward; e.g. (x a m a x).

Example in Haskell:

```
*Main> isPalindrome [1,2,3]
False
*Main> isPalindrome "madamimadam"
True
*Main> isPalindrome [1,2,4,8,16,8,4,2,1]
True
```

Solution:

```
isPalindrome :: (Eq a) => [a] -> Bool
isPalindrome xs = xs == (reverse xs)
```

## Problem 7

(**) Flatten a nested list structure.

Transform a list, possibly holding lists as elements into a `flat' list by replacing each list with its elements (recursively).

Example:

* (my-flatten '(a (b (c d) e))) (A B C D E)

Example in Haskell:

```
*Main> flatten (Elem 5)
[5]
*Main> flatten (List [Elem 1, List [Elem 2, List [Elem 3, Elem 4], Elem 5]])
[1,2,3,4,5]
*Main> flatten (List [])
[]
```

Solution:

```
data NestedList a = Elem a | List [NestedList a]
flatten :: NestedList a -> [a]
flatten (Elem x) = [x]
flatten (List x) = concatMap flatten x
```

We have to defined a new data type, because lists in Haskell are homogeneous. [1, [2, [3, 4], 5]] is a type error. Therefore, we must have a way of representing a list that may (or may not) be nested.

Our NestedList datatype is either a single element of some type (Elem a), or a list of NestedLists of the same type. (List [NestedList a]).

## Problem 8

(**) Eliminate consecutive duplicates of list elements.

If a list contains repeated elements they should be replaced with a single copy of the element. The order of the elements should not be changed.

Example: * (compress '(a a a a b c c a a d e e e e)) (A B C A D E) Example in Haskell: *Main> compress ['a','a','a','a','b','c','c','a','a','d','e','e','e','e'] ['a','b','c','a','d','e']

Solution:

```
compress :: Eq a => [a] -> [a]
compress = map head . group
```

We simply group equal values together (group), then take the head of each.
Note that (with GHC) we must give an explicit type to *compress* otherwise we get:

```
Ambiguous type variable `a' in the constraint:
`Eq a'
arising from use of `group'
Possible cause: the monomorphism restriction applied to the following:
compress :: [a] -> [a]
Probable fix: give these definition(s) an explicit type signature
or use -fno-monomorphism-restriction
```

We can circumvent the monomorphism restriction by writing *compress* this way (See: section 4.5.4 of the report):

```
compress xs = map head $ group xs
```

## Problem 9

(**) Pack consecutive duplicates of list elements into sublists. If a list contains repeated elements they should be placed in separate sublists.

Example: * (pack '(a a a a b c c a a d e e e e)) ((A A A A) (B) (C C) (A A) (D) (E E E E)) <example in lisp> Example in Haskell:

Solution:

```
group (x:xs) = let (first,rest) = span (==x) xs
in (x:first) : group rest
group [] = []
```

'group' is also in the Prelude, here's an implementation using 'span'.

## Problem 10

(*) Run-length encoding of a list. Use the result of problem P09 to implement the so-called run-length encoding data compression method. Consecutive duplicates of elements are encoded as lists (N E) where N is the number of duplicates of the element E.

Example:

* (encode '(a a a a b c c a a d e e e e)) ((4 A) (1 B) (2 C) (2 A) (1 D)(4 E))

Example in Haskell:

encode "aaaabccaadeeee" [(4,'a'),(1,'b'),(2,'c'),(2,'a'),(1,'d'),(4,'e')]

Solution:

```
encode xs = map (\x -> (length x,head x)) (group xs)
```

Or writing it Pointfree:

```
encode :: Eq a => [a] -> [(Int, a)]
encode = map (\x -> (length x, head x)) . group
```

Or (ab)using the "&&&" arrow operator for tuples:

```
encode :: Eq a => [a] -> [(Int, a)]
encode xs = map (length &&& head) $ group xs
```