LeetCode 409: Longest Palindrome Solution

Master LeetCode problem 409 (Longest Palindrome), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

409. Longest Palindrome

Easy

Hash Table String Greedy

Problem Explanation

Explanation:

To find the length of the longest palindrome that can be built with the given letters, we need to count the frequency of each letter in the input string. A palindrome can have an even count of each character, and at most one character can have an odd count (which will be placed in the middle of the palindrome). Hence, we can iterate over the counts of each character and calculate the length of the palindrome accordingly.

Initialize a variable length to store the length of the longest palindrome.
Iterate over the counts of each character in the input string.
For each character count, add the maximum even count (characterCount // 2 * 2) to the length.
If there is any character with an odd count, add 1 to the length (as it can be placed in the middle of the palindrome).
Return the final length.

Time Complexity: O(n) where n is the length of the input string s. Space Complexity: O(1) since we are using a constant amount of space to store the counts of characters.

Solution Code

class Solution {
    public int longestPalindrome(String s) {
        int[] charCounts = new int[128];
        for (char c : s.toCharArray()) {
            charCounts[c]++;
        }
        
        int length = 0;
        for (int count : charCounts) {
            length += count / 2 * 2;
        }
        
        if (length < s.length()) {
            length++; // Add 1 for a single character that can be placed in the middle
        }
        
        return length;
    }
}

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