# Finding the longest common substring of multiple strings in Haskell

The task is to find the longest common substring of a multitude of long strings. I'm trying to find the best algorithm that wouldn't use syntax trees/arrays (as I don't know anything about them yet).

The idea was to use binary search to find the length of the desired substring, instead of simply trying n, then (n - 1) etc.

My code looks like this:

-- find all the substrings of length n
ngrams :: Int -> String -> [String]
ngrams n s | n > length s = []
| otherwise    = take n s : ngrams n (drop 1 s)

-- find the longest common substring of multiple strings
longestCommonSubstring :: [String] -> String
longestCommonSubstring xs = go 0 $length (head xs) + 1 where -- find a substring of a given length n that is common to all strings commonSubstrings n = foldr1 intersect (map (ngrams n) xs) go l r -- if the binary search ended, pick one common substring | r - l == 1 = head$ commonSubstrings l
| otherwise
= case commonSubstrings m of
[] -> go l m    -- the length is too big, try a smaller one
_  -> go m r    -- try longer length to find longer substring
where
m = (l + r) div 2    -- the middle point


It runs at around 3s for my dataset (~100 strings of length ~1000), which seems slow to me. Is there any way to clean up and quicken the code? And is there a better way to approach this problem (apart from the syntax trees) in general?

intersect runs in quadratic time. Sets can use Ord information to speed that up.

-- find the longest common substring of multiple strings
longestCommonSubstring :: [String] -> String
longestCommonSubstring xs = go 0 $length (head xs) + 1 where -- find a substring of a given length n that is common to all strings commonSubstrings n = foldr1 S.intersection (map (S.fromList . ngrams n) xs) go l r -- if the binary search ended, pick one common substring | r - l == 1 = S.findMin$ commonSubstrings l
| otherwise
= if S.null \$ commonSubstrings m
then go l m    -- the length is too big, try a smaller one
else go m r    -- try longer length to find longer substring
where
m = (l + r) div 2    -- the middle point


For comparison, here's an implementation that skips the binary search part:

-- find the longest common substring of multiple strings
longestCommonSubstring :: [String] -> String
longestCommonSubstring = maximumBy (comparing length)
. foldr1 S.intersection . map (S.fromList . map inits . tails)

• Good insights. Your version (with sets and with binary search) runs 2s longer then the one with lists, though. – Sh4rP EYE Jun 12 '17 at 15:17