[Haskell-cafe] One-Off detection. Question about Space/Time complexity
Andrew Butterfield
Andrew.Butterfield at scss.tcd.ie
Thu Mar 23 08:52:10 UTC 2017
It can be done in one sweep down both strings together - look for the first difference,
and then test hypotheses regarding the 5 possible edits
oneOff :: String -> String -> Bool
oneOff [] [] = False
oneOff [] (_:rest) = null rest
oneOff (_:rest) [] = null rest
oneOff s1@(c1:s1') s2@(c2:s2')
| c1 == c2 = oneOff s1' s2' -- keep looking
| otherwise
= s1' == s2 -- s2 is s1 with c1 deleted or s1 is s2 with c1 inserted
|| s2' == s1 -- s1 is s2 with c2 deleted or s2 is s1 with c2 inserted
|| s1' == s2' -- c1 is s1 was replaced by c2 in s2, or v.v.
I guess the above should be O(1) in space - it's just a list crawl with tests
O(n) in the length of the shortest String
Cheers,
Andrew
> On 22 Mar 2017, at 18:48, David Turner <dct25-561bs at mythic-beasts.com> wrote:
>
> I think the space must be O(m+n) since you are using strict Text values. But even if you weren't, you're using s1 and s2 in more than one place so they will be shared, leading to them both being fully in memory at once.
>
> The minimum possible space complexity could well be O(1) if using String or lazy Text.
>
> The time complexity looks like O(min(m,n)), because that's the cost of `zip`/`filter`, not counting the cost of loading the two Texts into memory in the first place (which will be O(m+n)). If you used String, it'd be O(m+n) since that's the cost of the two calls to `length`. I think the minimum possible time complexity will be O(min(m,n)), and that could include the cost of loading the strings too.
>
> On 22 Mar 2017 18:32, "Michael Litchard" <litchard.michael at gmail.com <mailto:litchard.michael at gmail.com>> wrote:
> Problem spec from CareerCup <https://careercup.com/question?id=5092486548553728>.
>
> Given two strings, return boolean True/False if they are only one edit apart. Edit can be insert/delete/update of only one character in the string. Eg.
>
> -True
>
> xyz,xz
>
> xyz,xyk
>
> xy,xyz
>
> -False
>
> xyz,xyz
>
> xyz,xzy
>
> x,xyz
>
> module Strings.OneLetter where`
>
> import Prelude
> import qualified Data.Text as T`
>
> oneLetter :: T.Text -> T.Text -> Bool
> oneLetter s1 s2
> | s1 == s2 = False
> | max_l > (min_l + 1) = False
> | max_l == min_l = diff_size == 1
> | otherwise = diff_size == 0
> where
> length_s1 = T.length s1
> length_s2 = T.length s2
> max_l = max length_s1 length_s2
> min_l = min length_s1 length_s2
> diff_size = length $ filter (\(a,b) -> a /= b) zipped
> zipped = T.zip s1 s2`
> So, I used Text instead of String, hoping I could take advantage of fusion. I have the following questions and my initial attempt to answer them.
>
> What is the time complexity of oneLetter 0(m+n) where m is the length of s1 and n is the length of s2
>
> what is the space complexity of oneLetter? I'm thinking due to laziness it's O(1), only two Chars are in memory at any one time, or two Ints. But I'm hazy on why. If this is wrong, please articulate why. If I'm right, and my reasoning is wrong or incomplete, please say why.
>
> I don't think I can improve the time complexity. Am I right? Can the space complexity be improved?
>
> What if I changed from Text to String? I don't think the time complexity changes, but how does this change the space complexity?
>
>
>
>
>
>
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Andrew Butterfield
School of Computer Science & Statistics
Trinity College
Dublin 2, Ireland
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