2Max
June 5, 2020
One way to solve this problem is to collect input numbers with similar digit-sums in a hash table, compute the two-maximums of each hash set, and return the maximum. Instead, we solve the problem using the cluster function of a previous exercise and a few auxiliary functions:
(define (2max xs) (if ( xs))))
(define (sum-digits n) (sum (digits n)))
(define (f x)
(apply max (map sum (map 2max
(cluster sum-digits <lt; xs)))))
Here are two examples:
> (f '(51 17 71 42)) 93 > (f '(1 2 3 4 5 6)) -1
The pieces are individually small and easy to understand, which makes the whole function equally easy to understand. You can run the program at https://ideone.com/KcT9Ix.
Programming Praxis 
Interesting exercise. Here is my take on it using Julia 1.4: https://pastebin.com/uEPBeKBc
Have a nice weekend!
Scan the array, keeping track of the largest number seen so far for each digit sum:
def max2(a): s = {} # s[n] = largest k seen so far with dsum(k) = n max = -1 for n in a: dsum = sum(int(c) for c in str(n)) if dsum not in s: s[dsum] = n # A new digit sum else: m = s[dsum] if max < 0 or m+n > max: max = m+n # Update max if n > m: s[dsum] = n # Update s return maxHere’s a not-particularly efficient, not-particularly readable solution in Python.
def two_max(l): p = sorted(((sum(int(c) for c in str(x)), x) for x in l), reverse=True) return max((b + d for (a, b), (c, d) in zip(p, p[1:]) if a == c), default=-1) print(two_max([51, 71, 17, 42])) print(two_max([1, 2, 3, 4, 5, 6]))Output:
Here’s a Haskell version.
import Data.List (insertBy, unfoldr) import Data.Map (elems, fromListWith) import Data.Ord (Down(..), comparing) max2 :: Integral a => [a] -> a max2 xs = maximum ((-1) : top2Sums xs) top2Sums :: Integral a => [a] -> [a] top2Sums = map sum . filter ((==2) . length) . elems . fromListWith (\[n] -> top2 n) . digitSums top2 :: Ord a => a -> [a] -> [a] top2 x = take 2 . insertBy (comparing Down) x digitSums :: Integral a => [a] -> [(a, [a])] digitSums = map (\x -> (sum (digits x), [x])) digits :: Integral a => a -> [a] digits = unfoldr step where step 0 = Nothing step n = let (q, r) = quotRem n 10 in Just (r, q) main :: IO () main = mapM_ (print . max2) [ [] , [0] , [51, 71, 17, 42] , [51, 71, 17, 42, 2223] , [51, 71, 17, 42, 2222] ]