# Super Kth LIS

# Super Kth LIS

+ 0 comments **Here is Super Kth LIS problem solution in Python Java C++ and C programming**- https://programs.programmingoneonone.com/2021/07/hackerrank-super-kth-LIS-problem-solution.html

+ 0 comments Haskel Solution

{-# LANGUAGE ScopedTypeVariables #-} -- via editorial module Main where import qualified Data.Vector.Unboxed as U import qualified Data.Vector as V import qualified Data.Vector.Mutable as M import qualified Data.Vector.Unboxed.Mutable as UM import Data.List import Control.Monad.ST import Control.Monad.Primitive import Data.Bits import Data.Function import Data.Maybe main :: IO () main = do [_, k] <- fmap (map read . words) getLine :: IO [Int] as <- fmap (U.fromList . map read . words) getLine :: IO A let gs = stage1 as let rcs = stage2 as gs putStrLn (stage3 as gs rcs k) type A = UVec Int type G = [[Int]] type C = UVec Int type UVec = U.Vector type UMVec s a = UM.MVector s a stage1 :: A -> G stage1 as = runST $ do bs <- UM.replicate (succ n) 0 gs <- M.replicate (succ n) [] let f [] = return [] f ((i, a) : xs) = do l <- fmap (succ . maximum . (0:)) $ query bs (pred a) update bs (max l) a M.modify gs (i:) l fmap (l:) (f xs) ls <- f $ zip [0..] (U.toList as) fmap (take (maximum ls) . tail . V.toList) (V.freeze gs) where n = U.length as stage2 :: A -> G -> C stage2 as gs = runST $ do bs <- UM.replicate (succ n) 0 rcs <- UM.new n -- right counts sequence_ [UM.write rcs i 1 | i <- last gs] let handleGroup (g1:gs1) (g2:gs2) = do writeCounts g1 g2 sequence_ [clear bs (succ n - getA i) | i <- g2] handleGroup gs1 gs2 handleGroup _ _ = return () writeCounts (i:g1) g2 = do updateBs g3 rc <- fmap (foldl' add 0) $ query bs (succ n - succ (getA i)) UM.write rcs i rc writeCounts g1 g4 where (g3, g4) = span (> i) g2 writeCounts _ _ = return () updateBs [] = return () updateBs (i:g1) = do rc <- UM.read rcs i update bs (add rc) (succ n - getA i) updateBs g1 handleGroup (tail . reverse $ gs) (reverse gs) U.freeze rcs where getA = (as U.!) n = U.length as stage3 :: A -> G -> C -> Int -> String stage3 as gs rcs k = runST $ do bs <- UM.replicate (n + 2) 0 let go _ [] _ = return [] go k1 (dg:dgs1) ii = do mb <- findIi a_mb dg k1 case mb of Nothing -> return [] (Just (jj_lcs, k2)) -> do sequence_ [clear bs (succ i) | i <- ii] sequence_ [update bs (add lc) (succ j) | (j, lc) <- jj_lcs] let b = getA . fst . head $ jj_lcs fmap (b :) (go k2 dgs1 (map fst jj_lcs)) where a_mb = if null ii then Nothing else (Just (getA (head ii))) findIi _ [] _ = return Nothing findIi a_mb (ii:iis) k1 | isJust a_mb && b <= a = findIi a_mb iis k1 | otherwise = do lcs <- if isJust a_mb then sequence [fmap (foldl' add 0) (query bs (succ i)) | i <- ii] else return (repeat 1) -- left counts let k2 = foldl' add 0 [mult lc (getRCount i) | (lc, i) <- zip lcs ii] if k2 < k1 then findIi a_mb iis (k1 - k2) else return (Just (zip ii lcs, k1)) where b = getA (head ii) (Just a) = a_mb xs <- go k dgs [] return $ if null xs then "-1" else (intercalate " " (map show xs)) where n = U.length as -- divided groups (groups = indices grouped by left-length, divided = every group is again grouped, now by a-value. these subgroups we call inverse images) dgs = [groupBy ((==) `on` getA) . sortBy (compare `on` getA) $ g1 | g1 <- gs] getA = (as U.!) getRCount = (rcs U.!) -- lc = left count = number of ways to reach this position query :: PrimMonad m => UMVec (PrimState m) Int -> Int -> m [Int] query bs i = sequence [UM.read bs j | j <- takeWhile (> 0) (iterate (\x -> x - (x .&. (-x))) i)] update :: PrimMonad m => UMVec (PrimState m) Int -> (Int -> Int) -> Int -> m () update bs f i = sequence_ [UM.modify bs f j | j <- takeWhile (< UM.length bs) (iterate (\x -> x + (x .&. (-x))) i)] clear :: PrimMonad m => UMVec (PrimState m) Int -> Int -> m () clear bs i = sequence_ [UM.write bs j 0 | j <- takeWhile (< UM.length bs) (iterate (\x -> x + (x .&. (-x))) i)] kBound :: Int kBound = 10 ^ (18 :: Int) + 1 add :: Int -> Int -> Int add x y = min (x + y) kBound mult :: Int -> Int -> Int mult _ 0 = 0 mult x y = if x > 1 + (kBound `quot` y) then kBound else (x * y)

+ 0 comments minor point here but I keep getting a wrong score on case 16 (which does not show my output) -- running in visual studio I appear to be getting the correct result?

+ 1 comment testcase 8, can it be wrong:

the range 0 to 145, i have the same lis (the smallest prefix) as the solution. at index 146, the testcase expects to take 6353 which is the smallest, has only 648 children and thus covers rank 1..648, that is not enough for k=1000. I take the next higher which is 6538 and covers rank 649..1296. I have verified for hours but can't find the error (index are 0-based). I even see the path the expected result goes, because I have found that sequence too with a lower rank tough (between 217 and 648). I cross-checked the code several times.

Could anybody verify this.

here are some details about test case 8: - LIS length: 196 - # of LIS: 17'236'010'373'120 - sequence start: 38, 45, 96, 126, .. - sequence end: 9385, 9483, 9579, 9650, 9776, 9835, 9948

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