# Difference between revisions of "99 questions/Solutions/18"

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< 99 questions | Solutions

(another solution using splitAt) |
(I've come up with a solution that actually is an extended version of the 1st solution) |
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<haskell> |
<haskell> |
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slice xs (i+1) k = take (k-i) $ drop i xs |
slice xs (i+1) k = take (k-i) $ drop i xs |
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+ | </haskell> |
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+ | |||

+ | The same solution as above, but with guards: |
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+ | |||

+ | <haskell> |
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+ | slice [] _ _ = [] |
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+ | slice xs k n | k == n = [] |
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+ | | k > n = error "k > n" |
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+ | | k == 0 = take n xs |
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+ | | otherwise = drop (k-1) $ take n xs |
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</haskell> |
</haskell> |
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## Revision as of 13:00, 18 August 2010

(**) 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 with guards:

```
slice [] _ _ = []
slice xs k n | k == n = []
| k > n = error "k > n"
| k == 0 = take n xs
| otherwise = 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
```