Approach: Count freq — pairs + one odd center

def longestPalindrome(s):
    from collections import Counter
    length = 0; odd = False
    for v in Counter(s).values():
        length += (v // 2) * 2
        if v % 2: odd = True
    return length + (1 if odd else 0)

Complexity: O(n) time, O(1) space.

Full Solution — all languages

class Solution:
    def longestPalindrome(self, s):
        from collections import Counter
        length = 0
        odd = False
        for v in Counter(s).values():
            length += (v // 2) * 2
            if v % 2:
                odd = True
        return length + (1 if odd else 0)

import java.util.*;
class Solution {
    public int longestPalindrome(String s) {
        Map<Character,Integer> f = new HashMap<>();
        for (char c : s.toCharArray()) f.merge(c, 1, Integer::sum);
        int length = 0;
        boolean odd = false;
        for (int v : f.values()) {
            length += (v / 2) * 2;
            if (v % 2 == 1) odd = true;
        }
        return length + (odd ? 1 : 0);
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int longestPalindrome(string s) {
        unordered_map<char,int> f;
        for (char c : s) f[c]++;
        int length = 0;
        bool odd = false;
        for (auto& kv : f) {
            length += (kv.second / 2) * 2;
            if (kv.second % 2) odd = true;
        }
        return length + (odd ? 1 : 0);
    }
};

/**
 * @param {string} s
 * @return {number}
 */ var longestPalindrome = function(s) {
    const f = {};
    for (const c of s) f[c] = (f[c] || 0) + 1;
    let length = 0, odd = false;
    for (const k in f) {
        length += Math.floor(f[k] / 2) * 2;
        if (f[k] % 2) odd = true;
    }
    return length + (odd ? 1 : 0);
};

Longest Palindrome

Example 1

Example 2

Example 3

Constraints