DEV Community

Discussion on: Daily Coding Problem #1

View post

Replies for: I have come up with this sumPair :: Num a => (a, a) -> a sumPair (x, y) = x + y isMoreThanOnce :: Eq a => a -> [a] -> Bool isMore...

Collapse
 
caseywebb profile image
Casey Webb •

Here's another solution in Haskell

import qualified Data.Set as Set

solve :: Integral a => a -> [a] -> Bool
solve = solve' Set.empty

solve' :: Integral a => Set.Set a -> a -> [a] -> Bool
solve' _ _ [] = False
solve' possibleAddends targetSum (x:xs)
  | Set.member x possibleAddends = True
  | otherwise = solve' (Set.insert addend possibleAddends) targetSum xs
  where addend = targetSum - x