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Use dictionary comprehension and s.count()
s.count()
Map modulo two across the counts, odd = 1, even = 0, summing the result
If there is more than one odd result cannot be a palindrome
d = {c:s.count(c) for c in set(s)} if sum(map(lambda x: x%2, d.values())) > 1: return 'NO' return 'YES'
That looks a lot like the Haskell solution:
import Data.List(sort, group) main = interact isAnaPalin isAnaPalin s = if (length $ filter id $ map (odd . length) $ group $ sort s) > 1 then "NO" else "YES"
Game of Thrones - I
You are viewing a single comment's thread. Return to all comments →
Use dictionary comprehension and
s.count()Map modulo two across the counts, odd = 1, even = 0, summing the result
If there is more than one odd result cannot be a palindrome
That looks a lot like the Haskell solution: