# 99 questions/Solutions/18

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< 99 questions | Solutions

Revision as of 15:59, 18 August 2010 by Giu (talk | contribs) (Updated my solution to a more paranoid one)

(**) 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 k'th element of the original list (both limits included). Start counting the elements with 1.

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

The same solution as above, but the more paranoid (maybe too paranoid?) version of it (uses guards and Maybe):

```
slice :: [a] -> Int -> Int -> Maybe [a]
slice [] _ _ = Just []
slice xs k n | k == n = Just []
| k > n || k > length xs || n > length xs || k < 0 || n < 0 = Nothing
| k == 0 = Just (take n xs)
| otherwise = Just (drop (k-1) $ take n xs)
```

Or, an iterative solution:

```
slice :: [a]->Int->Int->[a]
slice lst 1 m = slice' lst m []
where
slice' :: [a]->Int->[a]->[a]
slice' _ 0 acc = reverse acc
slice' (x:xs) n acc = slice' xs (n - 1) (x:acc)
slice (x:xs) n m = slice xs (n - 1) (m - 1)
```

Or:

```
slice :: [a] -> Int -> Int -> [a]
slice (x:xs) i k
| i > 1 = slice xs (i - 1) (k - 1)
| k < 1 = []
| otherwise = x:slice xs (i - 1) (k - 1)
```

Another way using `splitAt`

, though not nearly as elegant as the `take`

and `drop`

version:

```
slice :: [a] -> Int -> Int -> [a]
slice xs i k = chunk
where chop = snd $ splitAt i' xs -- Get the piece starting at i
chunk = fst $ splitAt (k - i') chop -- Remove the part after k
i' = i - 1
```