💯Faster✅💯Less Mem✅🧠Detailed Approach🎯Set Approach🔥Python🐍Java☕C++😎
\uD83D\uDE80 Hi, I\'m Mohammed Raziullah Ansari, and I\'m excited to share solution to this question with detailed explanation:\n\n# \uD83C\uDFAFProblem Explain
# \uD83D\uDE80 Hi, I\'m [Mohammed Raziullah Ansari](https://leetcode.com/Mohammed_Raziullah_Ansari/), and I\'m excited to share solution to this question with detailed explanation:\n\n# \uD83C\uDFAFProblem Explaination: \nThe problem is to find the length of the longest palindrome that can be constructed using the characters from a given string. \n\nA palindrome is a sequence of characters that reads the same backward as forward. For example, "racecar" and "level" are palindromes.\n\n## \uD83D\uDCE5Input\n- A single string \\( s \\) containing only lowercase and/or uppercase English letters.\n\n## \uD83D\uDCE4Output\n- An integer representing the length of the longest possible palindrome that can be formed with the characters of the given string.\n\n# \uD83E\uDDE0Thinking Behind the Solution:\n\nThe idea behind the solution is to use a set to help us count pairs of characters, and then determine if we can have a single middle character in our palindrome. Here\'s the thought process:\n\n1. **Initialize a Set:**\n   - We start with an empty set. This set will help us keep track of characters that appear an odd number of times.\n\n2. **Iterate Over the Characters:**\n   - For each character in the string, we check if it is already in the set.\n\n3. **Check for Pairs:**\n   - If the character is in the set, it means we have found a pair (one character we saw earlier plus this one).\n   - We remove this character from the set because we have used up the pair.\n   - We increase our palindrome length by 2 because a pair of characters can be placed symmetrically in the palindrome (one on the left side and one on the right side).\n\n4. **Track Odd Characters:**\n   - If the character is not in the set, we add it to the set. This means we have seen this character once, and it might be part of a pair if we see it again later.\n\n5. **Check for Middle Character:**\n   - After processing all characters, if there are any characters left in the set, it means these characters appear an odd number of times.\n   - We can use one of these characters as the middle character of the palindrome (since a palindrome can have one character in the center without a pair).\n\n6. **Calculate the Length:**\n   - The total length of the palindrome will be the number of pairs we found (each pair contributes 2 to the length).\n   - If we have any characters left in the set, we add 1 to the length for the single middle character.\n\n# Let\'s walkthrough\uD83D\uDEB6\uD83C\uDFFB\u200D\u2642\uFE0F the implementation process with an example for better understanding\uD83C\uDFAF:\n\nLet\'s take the string "abccccdd" as an example:\n\n1. **Initial State:**\n   - Set: {}\n\n2. **Process Each Character:**\n   - \'a\' is not in the set. Add \'a\'. Set: {\'a\'}\n   - \'b\' is not in the set. Add \'b\'. Set: {\'a\', \'b\'}\n   - \'c\' is not in the set. Add \'c\'. Set: {\'a\', \'b\', \'c\'}\n   - Another \'c\'. It\'s already in the set. Remove \'c\'. Increase length by 2. Length: 2. Set: {\'a\', \'b\'}\n   - Another \'c\'. Add \'c\'. Set: {\'a\', \'b\', \'c\'}\n   - Another \'c\'. It\'s already in the set. Remove \'c\'. Increase length by 2. Length: 4. Set: {\'a\', \'b\'}\n   - \'d\' is not in the set. Add \'d\'. Set: {\'a\', \'b\', \'d\'}\n   - Another \'d\'. It\'s already in the set. Remove \'d\'. Increase length by 2. Length: 6. Set: {\'a\', \'b\'}\n\n3. **Final Check:**\n   - The set still has {\'a\', \'b\'}, meaning we have characters that appeared an odd number of times.\n   - We can use one of these characters as the center of the palindrome. Add 1 to the length. Final Length: 7.\n\nSo, the longest palindrome we can build with "abccccdd" has a length of 7.\n# \u2705Approach:\n1. **Initialize Variables:**\n   - Create an empty set to keep track of characters that appear an odd number of times.\n   - Initialize a variable `length` to keep track of the length of the palindrome.\n\n2. **Iterate Over the Characters of the String:**\n   - For each character in the string, check if it is in the set.\n   - If it is in the set, remove it from the set and increase the `length` by 2.\n   - If it is not in the set, add it to the set.\n\n3. **Check for Odd Characters:**\n   - After processing all characters, check if there are any characters left in the set.\n   - If the set is not empty, it means there are characters that appear an odd number of times. Add 1 to the `length` for the center character of the palindrome.\n\n4. **Return the Length:**\n   - Return the calculated `length` of the longest palindrome.\n\n# Code\uD83D\uDC69\uD83C\uDFFB\u200D\uD83D\uDCBB:\n```Python []\nclass Solution:\n    def longestPalindrome(self, s: str) -> int:\n        # Initialize a set to keep track of characters with odd frequencies\n        char_set = set()\n        # Initialize the length of the longest palindrome\n        length = 0\n        \n        # Iterate over each character in the string\n        for char in s:\n            # If the character is already in the set, remove it and increase the length by 2\n            if char in char_set:\n                char_set.remove(char)\n                length += 2\n            # If the character is not in the set, add it to the set\n            else:\n                char_set.add(char)\n        \n        # If there are any characters left in the set, add 1 to the length for the middle character\n        if char_set:\n            length += 1\n        \n        # Return the total length of the longest palindrome\n        return length\n```\n```Java []\npublic class Solution {\n    public int longestPalindrome(String s) {\n        // Initialize a set to keep track of characters with odd frequencies\n        HashSet<Character> charSet = new HashSet<>();\n        // Initialize the length of the longest palindrome\n        int length = 0;\n        \n        // Iterate over each character in the string\n        for (char c : s.toCharArray()) {\n            // If the character is already in the set, remove it and increase the length by 2\n            if (charSet.contains(c)) {\n                charSet.remove(c);\n                length += 2;\n            } else {\n                // If the character is not in the set, add it to the set\n                charSet.add(c);\n            }\n        }\n        \n        // If there are any characters left in the set, add 1 to the length for the middle character\n        if (!charSet.isEmpty()) {\n            length += 1;\n        }\n        \n        // Return the total length of the longest palindrome\n        return length;\n    }\n}\n\n```\n```C++ []\nclass Solution {\npublic:\n    int longestPalindrome(const std::string& s) {\n        // Initialize a set to keep track of characters with odd frequencies\n        std::unordered_set<char> charSet;\n        // Initialize the length of the longest palindrome\n        int length = 0;\n        \n        // Iterate over each character in the string\n        for (char c : s) {\n            // If the character is already in the set, remove it and increase the length by 2\n            if (charSet.find(c) != charSet.end()) {\n                charSet.erase(c);\n                length += 2;\n            } else {\n                // If the character is not in the set, add it to the set\n                charSet.insert(c);\n            }\n        }\n        \n        // If there are any characters left in the set, add 1 to the length for the middle character\n        if (!charSet.empty()) {\n            length += 1;\n        }\n        \n        // Return the total length of the longest palindrome\n        return length;\n    }\n};\n\n\n```\n\n# \uD83D\uDCA1 I invite you to check out [my profile](https://leetcode.com/Mohammed_Raziullah_Ansari/) for detailed explanations and code for each method. Happy coding and learning! \uD83D\uDCDA
['String', 'C++', 'Java', 'Python3']