I was writing code to check if any permutation of a string can make it palindrome and came up with the following logic:
anagram = [ x for x in input() ] count_alphabets = [ anagram.count(x) for x in set(anagram) ] middle_element = [x for x in count_alphabets if x % 2 == 1] print("YES") if len(middle_element) == 0 or len(middle_element) == 1 else print("NO")
The logic is to count the number of times each character appears. ONLY one or none of the characters can appear odd number of times.
The above logic works fine and passes all test cases. Is there a better logic for this?