A palindrome is a series of characters that read the same forwards as backwards such as "hannah", "racecar" and "lol".

For this Kata you need to write a function that takes a string of characters and returns the length, as an integer value, of longest alphanumeric palindrome that could be made by combining the characters in any order but using each character only once. The function should not be case sensitive.

For example if passed "Hannah" it should return 6 and if passed "aabbcc_yYx_" it should return 9 because one possible palindrome would be "abcyxycba".

Difficulty: EASY
Tags: String algorithms

Data structures
Skill types: Data structures

Starter code:
def longest_palindrome(s):
	

Solutions:
from collections import Counter

def longest_palindrome(s):
	c = Counter(filter(str.isalnum, s.lower()))
	return sum((v // 2 * 2 for v in c.values())) + any((v % 2 for v in c.values()))


from collections import Counter
import re

def longest_palindrome(s):
	(n, odds, c) = (0, 0, Counter(re.sub('\\W+|_', '', s.lower())))
	for v in c.values():
		(x, r) = divmod(v, 2)
		n += 2 * x
		odds += r
	return n + bool(odds)


from collections import Counter

def longest_palindrome(s):
	freq = Counter((c for c in s.lower() if c.isalnum()))
	(even, odd) = ([], [])
	for (k, v) in freq.items():
		if v % 2:
			odd.append(v - 1)
		else:
			even.append(v)
	return sum(even) + (sum(odd) + 1 if odd else 0)


def longest_palindrome(s):
	new_s = ''.join([x for x in s.lower() if x.isalnum()])
	k = [new_s.count(x) for x in set(new_s)]
	res = 0
	for x in k:
		if x % 2 == 0:
			res += x
		else:
			res += x - 1
	return res + 1 if any((x % 2 == 1 for x in k)) else res


from collections import Counter

def longest_palindrome(s):
	if s == '':
		return 0
	s = s.upper()
	x = Counter(s)
	len = 0
	flag = False
	for i in x:
		if i.isalnum():
			if x[i] % 2 == 0:
				len += x[i]
			else:
				flag = True
				len += x[i] - 1
	if flag == True:
		return len + 1
	return len


from collections import Counter
from string import ascii_letters, digits

def longest_palindrome(data):
	cnt = Counter([d for d in data.lower() if d in ascii_letters + digits])
	even = sum([n for n in list(cnt.values()) if n % 2 == 0])
	odd = sorted([n for n in list(cnt.values()) if n % 2 != 0])[::-1]
	if len(odd) == 0:
		ans = even
	if len(odd) == 1:
		ans = even + odd[0]
	if len(odd) > 1:
		ans = even + odd[0] + sum([n - 1 for n in odd[1:]])
	return ans


from collections import Counter

def longest_palindrome(string):
	(possibles, has_odd) = (0, False)
	for (char, c) in Counter(string.lower()).items():
		if not char.isalnum():
			continue
		if c & 1:
			possibles += c - 1
			has_odd = True
		else:
			possibles += c
	return possibles + (1 if has_odd else 0)


from collections import Counter

def longest_palindrome(s):
	(d, r) = (Counter([x for x in s.lower() if x.isalnum()]), 0)
	for x in d.values():
		r += x // 2 * 2 + (x % 2 >> r % 2)
	return r


from collections import Counter

def longest_palindrome(s):
	flag = res = 0
	for v in Counter(map(str.lower, filter(str.isalnum, s))).values():
		flag |= v & 1
		res += v >> 1 << 1
	return res + flag


import re

def longest_palindrome(s):
	s = list(re.sub('[\\W_]+', '', s.lower()))
	solution = 0
	j = 0
	letter_count = 0
	for letter in s:
		letter_count = s.count(letter)
		if letter_count % 2 == 0:
			solution += letter_count
		else:
			solution += letter_count - 1
			j = 1
		s = [i for i in s if i != letter]
	return solution + j