Problem 9

Convert a list to a list of lists where repeated elements of the source list are packed into sublists. Elements that are not repeated should be placed in a one element sublist.

Example

pack [1,1,1,2,3,3,3,4,4,4,4,5,6,6] ==
    [ [1,1,1]
    , [2]
    , [3, 3, 3]
    , [4, 4, 4, 4]
    , [5]
    , [6, 6]
    ]

Unit Test

import Html
import List
import Maybe


pack : List a -> List (List a)
pack xs =
    -- your implementation goes here
    [ [] ]


main : Html.Html a
main =
    Html.text
        <| case test of
            0 ->
                "Your implementation passed all tests."

            1 ->
                "Your implementation failed one test."

            x ->
                "Your implementation failed " ++ (toString x) ++ " tests."


test : Int
test =
    List.length
        <| List.filter ((==) False)
            [ pack [ 1, 1, 1, 1, 2, 5, 5, 2, 1 ] == [ [ 1, 1, 1, 1 ], [ 2 ], [ 5, 5 ], [ 2 ], [ 1 ] ]
            , pack [ 2, 1, 1, 1 ] == [ [ 2 ], [ 1, 1, 1 ] ]
            , pack [ 2, 2, 2, 1, 1, 1 ] == [ [ 2, 2, 2 ], [ 1, 1, 1 ] ]
            , pack [ 1 ] == [ [ 1 ] ]
            , pack [] == []
            , pack [ "aa", "aa", "aa" ] == [ [ "aa", "aa", "aa" ] ]
            , pack [ "aab", "b", "b", "aa" ] == [ [ "aab" ], [ "b", "b" ], [ "aa" ] ]
            ]

Hints

1. One solution recurses using a helper function accumulating the output lists.
2. takeWhile and dropWhile could come in handy in another recursive solution.
3. A solution can be written using foldr. The function passed to foldr needs to return the same output as pack, namely List (List a).

Solutions

Solutions