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Maximize Palindrome Length From Subsequences

Difficulty: Hard

Problem Description

You are given two strings, word1 and word2. You want to construct a string by selecting some non-empty subsequence from each of the given strings and concatenating them. The goal is to determine the length of the longest palindrome that can be created through this method. If no palindromes can be constructed, return 0.

Key Insights

  • A palindrome reads the same forwards and backwards, meaning the first half must mirror the second half.
  • We can utilize the frequency of characters in both strings to determine potential pairs that can contribute to a palindrome.
  • Single characters can also contribute to palindrome lengths if they appear in both strings.

Space and Time Complexity

Time Complexity: O(1) - The character frequency calculation is constant time since there are only 26 lowercase letters. Space Complexity: O(1) - We only require a fixed amount of space for character counts.

Solution

To solve the problem, we will:

  1. Count the frequency of each character in both word1 and word2.
  2. For each character, calculate pairs that can be formed from both strings.
  3. Sum these pairs to form the base length of the palindrome.
  4. Check if there's any character that can be placed in the middle if the palindrome length is even, thus allowing for an additional character.

This approach primarily uses arrays to store character counts and implements a greedy strategy to maximize the palindrome length.

Code Solutions

def maxPalindromeLength(word1, word2):
    from collections import Counter
    
    # Count frequency of characters in both words
    count1 = Counter(word1)
    count2 = Counter(word2)
    palindrome_length = 0
    odd_found = False

    # Iterate over each character in the alphabet
    for char in 'abcdefghijklmnopqrstuvwxyz':
        if char in count1 and char in count2:
            # Add pairs from both counts
            pairs = min(count1[char], count2[char])
            palindrome_length += pairs * 2
            
            # Check for odd counts
            if pairs > 0:
                odd_found = True

    # If there's at least one character that can be a middle character
    if odd_found:
        palindrome_length += 1
    
    return palindrome_length
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