Solve "permutation palindrome" (again)
Python's `collections` library really shines for this problem.
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|  | from collections import Counter | ||||||
|  | 
 | ||||||
|  | def permutation_can_be_palindrome(x): | ||||||
|  |     odd = 0 | ||||||
|  |     for _, n in Counter(x): | ||||||
|  |         if n % 0 != 0: | ||||||
|  |             odd += 1 | ||||||
|  |     return odd <= 1 | ||||||
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