# Difference between revisions of "99 questions/Solutions/19"

(Added a new solution using recursive function composition.) |
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</haskell> |
</haskell> |
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+ | Here's another solution without using length. If the order of the arguments is reversed so that the integer comes before the list, then it is possible to define rotation by n as rotation by (n-1) followed by rotation by 1 recursively (with suitable modifications to support negative rotations). This leads to the following solution: |
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+ | <haskell> |
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+ | rotate :: [a] -> Int -> [a] |
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+ | rotate xs n = rot n xs |
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+ | where rot 0 = id |
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+ | rot 1 = \xs -> case xs of [] -> []; xs -> tail xs ++ [head xs] |
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+ | rot (-1) = \xs -> case xs of [] -> []; xs -> (last xs):init xs |
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+ | rot n |
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+ | | n > 0 = (rot (n-1)).(rot 1) |
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+ | | n < 0 = (rot (n+1)).(rot (-1)) |
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+ | </haskell> |
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[[Category:Programming exercise spoilers]] |
[[Category:Programming exercise spoilers]] |

## Revision as of 18:18, 13 June 2020

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

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

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

(Note that this solution uses n+k-patterns which are removed from Haskell 2010.)

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 using list comprehension (only works for sequential increasing elements):

```
rotate :: (Enum a) => [a] -> Int -> [a]
rotate xs n = [(f n) .. last xs] ++ [head xs .. (f (n-1))]
where f k = xs !! (k `mod` length xs)
```

or without mod:

```
rotate xs n = take (length xs) $ drop (length xs + n) $ cycle xs
```

or

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

or

```
rotate xs n | n >= 0 = drop n xs ++ take n xs
| n < 0 = drop len xs ++ take len xs
where len = n+length xs
```

or calculate the position at first:

```
rotate xs n = let i = if n < 0 then length xs + n else n
in drop i xs ++ take i xs
```

or

```
rotate xs n = drop nn xs ++ take nn xs
where
nn = n `mod` length xs
```

Using a simple splitAt trick

```
rotate xs n
| n < 0 = rotate xs (n+len)
| n > len = rotate xs (n-len)
| otherwise = let (f,s) = splitAt n xs in s ++ f
where len = length xs
```

Without using `length`

:

```
rotate xs n
| n > 0 = (reverse . take n . reverse $ xs) ++ (reverse . drop n . reverse $ xs)
| n <= 0 = (drop (negate n) xs) ++ (take (negate n) xs)
```

A much simpler solution without using `length`

that is very similar to the first solution:

```
rotate :: [a] -> Int -> [a]
rotate [] _ = []
rotate x 0 = x
rotate x y
| y > 0 = rotate (tail x ++ [head x]) (y-1)
| otherwise = rotate (last x : init x) (y+1)
```

Here's another solution without using length. If the order of the arguments is reversed so that the integer comes before the list, then it is possible to define rotation by n as rotation by (n-1) followed by rotation by 1 recursively (with suitable modifications to support negative rotations). This leads to the following solution:

```
rotate :: [a] -> Int -> [a]
rotate xs n = rot n xs
where rot 0 = id
rot 1 = \xs -> case xs of [] -> []; xs -> tail xs ++ [head xs]
rot (-1) = \xs -> case xs of [] -> []; xs -> (last xs):init xs
rot n
| n > 0 = (rot (n-1)).(rot 1)
| n < 0 = (rot (n+1)).(rot (-1))
```