from the letters of that word. A letter can be used at most as many times as it appears in the input word. So, "letter" can only match words with 0, 1, or 2 t's in them.
frequencies = map (\x -> (head x, length x)) . group . sort superset xs = \ys -> let y = frequencies ys in length y == lx && and (zipWith (\(c,i) (d,j) -> c == d && i >= j) x y) where x = frequencies xs lx = length x
As far as I understand the spec, this algorithm is not correct: superset "ubuntu" "tun" == False Is at least one 'b' necessary, yes or no? If the answer is no, the following algorithm solves the problem and is faster then the one above: del y = del_acc [] where del_acc _ [] = mzero del_acc v (x:xs) | x == y = return (v++xs) del_acc v (x:xs) = del_acc (x:v) xs super u = not . null . foldM (flip del) u main = interact $ unlines . filter ("ubuntu" `super`) . lines BR, -- -- Mirko Rahn -- Tel +49-721 608 7504 -- --- http://liinwww.ira.uka.de/~rahn/ ---