🚀 100% faster approach, Using Hash Table, BRUTE FORCE
Intuition\n Describe your first thoughts on how to solve this problem. \nIf you see examples, and do a dry run, for each random string, you might see a pattern,
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIf you see examples, and do a dry run, for each random string, you might see a pattern, which is:\nWhen we have to create a palindrome string, we need **EVEN NUMBER OF CHARACTERS + 1**, so for that, if we just count each character frequency, and check some conditions,\n\n**EXMAPLE**: \nLets, say for a **string s** = "aabbbccdccdddd";\n\nThe frequency will be:\nmap:\na -> 2\nb -> 3\nc -> 4\nd -> 5\n\nAnd now, we just have to add all even number of frequency, and for odd numbers, we just have to ignore 1 and add all odd numbers with - 1;\n\nAnd if we do so, we\'ll get the maximum length of palindrome we can make using input string;\n\nCOUNTER = 0;\n\nEVENS:\na -> 2\nc -> 4\n\nCOUNTER = 2 + 4 = 6;\n\nODDS:\nb -> 3\nd -> 5\n\nCOUNTER = 6 + (5-1) + (3-1) = 12\n\n**AND, RETURN COUNTER + 1;** \n\n\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nCreate a hashmap, and store each character frequency, and sum up each even number of frequency, and odd number of frequency with a, - 1 and by ignoring 1\'s. And in the end return Counter + 1 for certain conditions, otherwise Counter. \n\n# Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(N) or O(52) MAX;\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int longestPalindrome(string s) {\n        // SUGGEST SOME IMPROVEMENTS IN COMMENTS\n        int n = s.size(), cnt = 0;\n        if(n==1 || n==0) return n;\n        if(n==2) {\n            if(s[0] == s[1]) return 2;\n            return 1;\n        } \n\n        unordered_map<char, int> mp;\n\n        for(auto i:s) {\n            ++mp[i];\n        }\n\n        bool evenExist = false, oneExist = false, oddExist = false;\n\n        for(auto i:mp) {\n            // IF one exist, that means, we have to ignore it for now\n            if(i.second == 1) {\n                oneExist = true;\n            }\n\n            // IF EVEN EXIST, just add it to count;\n            else if(i.second%2==0) {\n                evenExist = true;\n                cnt += i.second;\n            }\n\n            // If ODD EXIST, and its greater than 2, just add it to count with - 1;\n            else if(i.second > 2 && i.second%2 != 0) {\n                cnt += i.second-1;\n                oddExist = true;\n            }\n\n        }\n\n        // IF EVERY or IF ODD NOT EXIST or If EVEN NOT EXIST or IF NOT ONE EXIST or IF ONLY ODD EXIST\n        if(oddExist && evenExist && oneExist || \n            !oddExist && evenExist && oneExist || \n            oddExist && !evenExist && oneExist ||\n            oddExist && evenExist && !oneExist ||\n            oddExist && !evenExist && !oneExist) {\n            return cnt+1;\n        }\n        \n        else return cnt;\n    }\n};\n```
['Hash Table', 'Greedy', 'C++']