{x}
blog image

Maximize Palindrome Length From Subsequences

You are given two strings, word1 and word2. You want to construct a string in the following manner:

  • Choose some non-empty subsequence subsequence1 from word1.
  • Choose some non-empty subsequence subsequence2 from word2.
  • Concatenate the subsequences: subsequence1 + subsequence2, to make the string.

Return the length of the longest palindrome that can be constructed in the described manner. If no palindromes can be constructed, return 0.

A subsequence of a string s is a string that can be made by deleting some (possibly none) characters from s without changing the order of the remaining characters.

A palindrome is a string that reads the same forward as well as backward.

 

Example 1:

Input: word1 = "cacb", word2 = "cbba"
Output: 5
Explanation: Choose "ab" from word1 and "cba" from word2 to make "abcba", which is a palindrome.

Example 2:

Input: word1 = "ab", word2 = "ab"
Output: 3
Explanation: Choose "ab" from word1 and "a" from word2 to make "aba", which is a palindrome.

Example 3:

Input: word1 = "aa", word2 = "bb"
Output: 0
Explanation: You cannot construct a palindrome from the described method, so return 0.

 

Constraints:

  • 1 <= word1.length, word2.length <= 1000
  • word1 and word2 consist of lowercase English letters.

Solution Explanation: Maximize Palindrome Length From Subsequences

This problem asks to find the length of the longest palindrome that can be created by concatenating non-empty subsequences from two given strings, word1 and word2. The solution uses dynamic programming to efficiently solve this.

Approach:

  1. Concatenation: The first step is to concatenate word1 and word2 into a single string s. This simplifies the problem to finding the longest palindromic subsequence within s, with the added constraint that the palindrome must contain characters from both word1 and word2.

  2. Dynamic Programming (DP): A 2D DP table f is used, where f[i][j] represents the length of the longest palindromic subsequence within the substring s[i...j].

  3. Base Cases: The diagonal of the DP table (f[i][i]) is initialized to 1, as a single character is a palindrome of length 1.

  4. Iteration: The DP table is filled iteratively, starting from the bottom-right and moving towards the top-left. For each pair of indices (i, j):

    • If s[i] == s[j]: This means we have a potential palindrome extension. f[i][j] is set to f[i+1][j-1] + 2, extending the palindrome by two characters. A check is performed to ensure that s[i] and s[j] are from different input strings (word1 and word2). If they are, the global maximum length ans is updated.

    • If s[i] != s[j]: s[i] and s[j] cannot be part of the same palindrome. In this case, f[i][j] is the maximum of f[i+1][j] (excluding s[i]) and f[i][j-1] (excluding s[j]).

  5. Result: After filling the DP table, the value of ans represents the length of the longest palindrome that can be constructed according to the problem's conditions.

Time Complexity: O(n^2), where n is the length of the concatenated string s. This is due to the nested loops iterating through all pairs of indices in the DP table.

Space Complexity: O(n^2), the space used by the DP table f.

Code Examples (Python):

class Solution:
    def longestPalindrome(self, word1: str, word2: str) -> int:
        s = word1 + word2
        n = len(s)
        f = [[0] * n for _ in range(n)]
        for i in range(n):
            f[i][i] = 1
        ans = 0
        for i in range(n - 2, -1, -1):
            for j in range(i + 1, n):
                if s[i] == s[j]:
                    f[i][j] = f[i + 1][j - 1] + 2
                    if i < len(word1) <= j or j < len(word1) <= i: #Added condition for clarity
                        ans = max(ans, f[i][j])
                else:
                    f[i][j] = max(f[i + 1][j], f[i][j - 1])
        return ans

The code in other languages (Java, C++, Go, TypeScript, Rust) follows the same logic and algorithmic structure, only differing in syntax and specific language features. The provided code snippets in the original response demonstrate these variations effectively.