Longest Palindrome by Concatenating Two Letter Words

class Solution {
public:
    int maxPalindromeLength(vector<string>& wordSet) {
        const int sizeOfAlphabet = 26;
        vector matrix(sizeOfAlphabet, vector<int>(sizeOfAlphabet));
        for (const string& word : wordSet) {
            matrix[word[0] - 'a'][word[1] - 'a']++;
        }
        int result = 0;
        bool hasOddMiddle = false;
        for (int i = 0; i < sizeOfAlphabet; i++) {
            if (matrix[i][i] % 2 == 0) {
                result += matrix[i][i];
            } else {
                result += matrix[i][i] - 1;
                hasOddMiddle = true;
            }
            for (int j = i + 1; j < sizeOfAlphabet; j++) {
                result += 2 * min(matrix[i][j], matrix[j][i]);
            }
        }
        if (hasOddMiddle) {
            result++;
        }
        return 2 * result;
    }
};

In the provided C++ code, the goal is to determine the maximum length of a palindrome that can be formed by concatenating two-letter words from a given vector of strings, wordSet. The solution involves creating a 2D matrix to count occurrences of each two-letter combination, then using these counts to construct the longest possible palindrome.

• Matrix Initialization: The matrix dimensions are 26x26, representing all possible combinations of characters from 'a' to 'z'. This matrix is used to count the occurrences of each two-letter word in the given wordSet.

• Populating the Matrix: Each word in wordSet is broken into its constituent characters. The ASCII value of 'a' is subtracted from each character to convert them into an index from 0 to 25. These indices are used to increment the count in the matrix accordingly.

• Calculating the Result: The variable result is initialized to zero. A separate flag, hasOddMiddle, checks whether there's a central two-letter word with an odd count, allowing it to be the middle of the palindrome. For each entry in the matrix:

  • If a word with identical letters (like "aa" or "bb") is encountered and it has an even count, the entire count contributes to the palindrome length. If it's odd, all except one of those instances are used, and hasOddMiddle is set.
  • For two distinct letters (like "ab" and "ba"), the minimum of the counts of each pairing is taken and doubled (since each can serve as a mirror in the palindrome), and added to the result.

• Final Adjustments: If any word can be used in the middle of the palindrome (hasOddMiddle is true), one more instance is added to the result. Finally, the total palindrome length is doubled to account for both halves.

The approach efficiently calculates the maximum palindrome length using a combination of character indexing and matrix manipulation. This method optimizes the problem by reducing it from potential word combination checks to simple arithmetic operations based on frequencies, making it both time and space-efficient.

class Solution {
    public int maxPalindromeLength(String[] wordSet) {
        final int LETTERS = 26;
        int[][] pairs = new int[LETTERS][LETTERS];
        for (String word : wordSet) {
            pairs[word.charAt(0) - 'a'][word.charAt(1) - 'a']++;
        }
        int maxLength = 0;
        boolean hasMiddle = false;
        for (int i = 0; i < LETTERS; i++) {
            if (pairs[i][i] % 2 == 0) {
                maxLength += pairs[i][i];
            } else {
                maxLength += pairs[i][i] - 1;
                hasMiddle = true;
            }
            for (int j = i + 1; j < LETTERS; j++) {
                maxLength += 2 * Math.min(pairs[i][j], pairs[j][i]);
            }
        }
        if (hasMiddle) {
            maxLength += 1;
        }
        return 2 * maxLength;
    }
}

The problem requires creating the longest palindrome possible by concatenating two-letter words. The solution is implemented in Java, where the main steps are:

• Represent letter pair frequencies in a 2D array.
• Iterate through each pair, updating the palindrome's length using specific conditions.

Explanation of the Solution:

1. Initialize a 2D array (pairs) to store counts of each possible two-letter combination using their ASCII differences.

2. Fill the array with counts by looping through each word in the input wordSet.

3. Determine the maximum length of the palindrome:

   • For symmetric pairs (same first and second letter), add all pairs directly if their count is even. If odd, add the count minus one to make it even, and set a flag (hasMiddle) which indicates that an extra middle character can be added later.
   • For asymmetric pairs (different first and second letters), add the minimum of the pair and its reversal pair to the length. Multiply by 2 because each pair contributes to both halves of the palindrome.

4. After processing pairs, if hasMiddle is true, increase maxLength by 1 to include a central character.

5. Finally, return 2 * maxLength as each palindrome's half contributes twice when combined to form the full palindrome.

This offers an efficient solution, leveraging character pair counting and considering both symmetry and the central character for palindrome formation. Make sure to check for edge cases, such as no possible palindrome or single character palindromes, to fully adapt this method.

class Solution:
    def findLongestPalindromeLength(self, words: List[str]) -> int:
        size = 26
        pair_count = [[0 for _ in range(size)] for _ in range(size)]
        for word in words:
            pair_count[ord(word[0]) - ord('a')][ord(word[1]) - ord('a')] += 1
        total_length = 0
        has_center = False
        for i in range(size):
            if pair_count[i][i] % 2 == 0:
                total_length += pair_count[i][i]
            else:
                total_length += pair_count[i][i] - 1
                has_center = True
            for j in range(i + 1, size):
                total_length += 2 * min(pair_count[i][j], pair_count[j][i])
        if has_center:
            total_length += 1
        return 2 * total_length

This Python solution addresses the problem of finding the longest palindrome length that can be formed by concatenating two-letter words. The approach utilizes a 2D list pair_count to track the frequency of all possible 2-letter combinations derived from the input list of words.

• Initialize a 2D list pair_count of size 26x26 with zeros. This list represents letter combinations, where the rows and columns correspond to letters from 'a' to 'z'.

• Iterate through each word in the input list, updating pair_count to reflect the occurrences of each 2-letter word.

• Establish variables total_length to accumulate the length of possible palindromes and has_center to check if a palindrome can have a central character.

• Iterate through the grid:

  • Accumulate the pair counts for symmetric pairs (e.g., 'aa', 'bb') directly to total_length. If an uneven number exists in symmetric pairs, set aside one to potentially use as the central character in the palindrome.
  • For non-symmetric pairs (where the first and second letters differ), add the minimum occurrence between the pair and its reverse (e.g., for 'ab' and 'ba') to total_length.

• If a center character is feasible (has_center is true), increase total_length by one to include this character in the middle of the palindrome.

• Multiply the total_length by two to account for the mirrored structure of palindromes and return it as the final result.

This algorithm leverages the count of repeating two-letter words and smartly utilizes the minimum pairs to create the longest mirrored palindrome possible, ensuring an optimized and effective solution.