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

d = {c:s.count(c) for c in set(s)}
if sum(map(lambda x: x%2, d.values())) > 1:
    return 'NO'
return 'YES'