# 99 questions/11 to 20

### From HaskellWiki

Line 264: | Line 264: | ||

<haskell> | <haskell> | ||

− | rotate | + | rotate s n = if n >= 0 then |

drop n s ++ take n s | drop n s ++ take n s | ||

else let l = ((length s) + n) in | else let l = ((length s) + n) in |

## Revision as of 05:58, 5 January 2007

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

## 1 Problem 11

(*) Modified run-length encoding. Modify the result of problem 10 in such a way that if an element has no duplicates it is simply copied into the result list. Only elements with duplicates are transferred as (N E) lists.

Example: * (encode-modified '(a a a a b c c a a d e e e e)) ((4 A) B (2 C) (2 A) D (4 E)) Example in Haskell: P11> encodeModified "aaaabccaadeeee" [Multiple 4 'a',Single 'b',Multiple 2 'c',Multiple 2 'a',Single 'd',Multiple 4 'e']

Solution:

data ListItem a = Single a | Multiple Int a deriving (Show) encodeModified :: Eq a => [a] -> [ListItem a] encodeModified = map encodeHelper . encode where encodeHelper (1,x) = Single x encodeHelper (n,x) = Multiple n x

The ListItem definition contains 'deriving (Show)' so that we can get interactive output.

## 2 Problem 12

(**) Decode a run-length encoded list. Given a run-length code list generated as specified in problem 11. Construct its uncompressed version.

Example in Haskell: P12> decodeModified [Multiple 4 'a',Single 'b',Multiple 2 'c',Multiple 2 'a',Single 'd',Multiple 4 'e'] "aaaabccaadeeee"

Solution:

decodeModified :: [ListItem a] -> [a] decodeModified = concatMap decodeHelper where decodeHelper (Single x) = [x] decodeHelper (Multiple n x) = replicate n x

We only need to map single instances of an element to a list containing only one element and multiple ones to a list containing the specified number of elements and concatenate these lists.

## 3 Problem 13

(**) Run-length encoding of a list (direct solution). Implement the so-called run-length encoding data compression method directly. I.e. don't explicitly create the sublists containing the duplicates, as in problem 9, but only count them. As in problem P11, simplify the result list by replacing the singleton lists (1 X) by X.

Example: * (encode-direct '(a a a a b c c a a d e e e e)) ((4 A) B (2 C) (2 A) D (4 E)) Example in Haskell: P13> encodeDirect "aaaabccaadeeee" [Multiple 4 'a',Single 'b',Multiple 2 'c',Multiple 2 'a',Single 'd',Multiple 4 'e']

Solution:

encode' :: Eq a => [a] -> [(Int,a)] encode' = foldr helper [] where helper x [] = [(1,x)] helper x (y:ys) | x == snd y = (1+fst y,x):ys | otherwise = (1,x):y:ys encodeDirect :: Eq a => [a] -> [ListItem a] encodeDirect = map encodeHelper . encode' where encodeHelper (1,x) = Single x encodeHelper (n,x) = Multiple n x

## 4 Problem 14

(*) Duplicate the elements of a list.

Example: * (dupli '(a b c c d)) (A A B B C C C C D D) Example in Haskell: > dupli [1, 2, 3] [1,1,2,2,3,3]

Solution:

dupli [] = [] dupli (x:xs) = x:x:dupli xs

or, using list comprehension syntax:

concat [[x,x]|x<-[1,2,3]]

or, using the list monad:

dupli xs = xs >>= (\x -> [x,x])

## 5 Problem 15

(**) Replicate the elements of a list a given number of times.

Example: * (repli '(a b c) 3) (A A A B B B C C C) Example in Haskell: > repli "abc" 3 "aaabbbccc"

Solution:

repli :: [a] -> Int -> [a] repli xs n = concatMap (replicate n) xs

## 6 Problem 16

(**) Drop every N'th element from a list.

Example: * (drop '(a b c d e f g h i k) 3) (A B D E G H K) Example in Haskell: *Main> drop = "abcdefghik" 3 "abdeghk"

Solution:

drop xs n = drops xs (n-1) n drops [] _ _ = [] drops (x:xs) 0 max = drops xs (max-1) max drops (x:xs) (n+1) max = x:drops xs n max

Here, drops is a helper-function to drop. In drops, there is an index n that counts from max-1 down to 0, and removes the head element each time it hits 0.

Note that drop is one of the standard Haskell functions, so redefining it is generally not a good idea.

or using zip:

drop n = map snd . filter ((n/=) . fst) . zip (cycle [1..n])

## 7 Problem 17

(*) Split a list into two parts; the length of the first part is given.

Do not use any predefined predicates.

Example: * (split '(a b c d e f g h i k) 3) ( (A B C) (D E F G H I K)) Example in Haskell: *Main> split "abcdefghik" 3 ("abc", "defghik")

Solution using take and drop:

split xs n = (take n xs, drop n xs)

## 8 Problem 18

(**) Extract a slice from a list.

Given two indices, i and k, the slice is the list containing the elements between the i'th and i'th element of the original list (both limits included). Start counting the elements with 1.

Example: * (slice '(a b c d e f g h i k) 3 7) (C D E F G) Example in Haskell: *Main> slice ['a','b','c','d','e','f','g','h','i','k'] 3 7 "cdefg"

Solution:

slice xs (i+1) k = take (k-i) $ drop i xs

or

slice xs i j = [xs!!(i-1)..xs!!(j-1)]

## 9 Problem 19

(**) Rotate a list N places to the left.

Hint: Use the predefined functions length and (++).

Examples: * (rotate '(a b c d e f g h) 3) (D E F G H A B C) * (rotate '(a b c d e f g h) -2) (G H A B C D E F) Examples in Haskell: *Main> rotate ['a','b','c','d','e','f','g','h'] 3 "defghabc" *Main> rotate ['a','b','c','d','e','f','g','h'] (-2) "ghabcdef"

Solution:

rotate [] _ = [] rotate l 0 = l rotate (x:xs) (n+1) = rotate (xs ++ [x]) n rotate l n = rotate l (length l + n)

There are two separate cases:

- If n > 0, move the first element to the end of the list n times.
- If n < 0, convert the problem to the equivalent problem for n > 0 by adding the list's length to n.

or using cycle:

rotate xs n = take len . drop (n `mod` len) . cycle $ xs where len = length xs

or

rotate s n = if n >= 0 then drop n s ++ take n s else let l = ((length s) + n) in drop l s ++ take l s

## 10 Problem 20

(*) Remove the K'th element from a list.

Example in Prolog:

?- remove_at(X,[a,b,c,d],2,R). X = b R = [a,c,d]

Example in Lisp:

* (remove-at '(a b c d) 2) (A C D)

(Note that this only returns the residue list, while the Prolog version also returns the deleted element.)

Example in Haskell:

*Main> removeAt 1 removeAt 1 "abcd" ('b',"acd")

Solution:

removeAt :: Int -> [a] -> (a, [a]) removeAt k xs = case back of [] -> error "removeAt: index too large" x:rest -> (x, front ++ rest) where (front, back) = splitAt k xs

If the original list has fewer than k+1 elements, the second list will be empty, and there will be no element to extract. Note that the Prolog and Lisp versions treat 1 as the first element in the list, and the Lisp version appends NIL elements to the end of the list if k is greater than the list length.