class Solution:
    def longestPalindrome(self, s: str) -> int:      
        length = 0

        # count all the chars
        ct = {}
        for c in s:
            ct[c] = ct.get(c, 0) + 1

        # flag counter to see how many odd counts detected
        odd_detected = 0

        # traverse the set of chars
        for char in set(s):
    
            if ct[char] % 2 == 0:
                # even count, increment length
                length += ct[char]
            elif ct[char] % 2 == 1 and ct[char] > 2:
                # odd count is > 2, tally up as if it was odd count, but increment the flag
                # needed when s = 'xxxyyy'
                length += ct[char] - 1
                odd_detected += 1
            else:
                # odd count == 1, increment flag
                # needed when s = 'abcde'
                odd_detected += 1

        # check if any odds found, and increment length by one
        if odd_detected > 0:
            length += 1

        return length