Problem Description

In this problem, we are tasked with creating the longest palindrome possible by concatenating a subsequence from word1 with a subsequence from word2. A subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. A palindrome is a word or phrase that reads the same backwards as forwards.

The goal is to figure out the length of the longest palindrome that can be formed in this way. If it's not possible to create any palindrome, the result should be 0. It is important to note that subsequence1 and subsequence2 must both be non-empty.

Intuition

To arrive at the solution, we have to consider that a palindrome reads the same forwards and backwards. Therefore, we need to look for matching characters in word1 and word2. Since we can form subsequences, we are not restricted to contiguous characters and we can skip some in order to maximize the length of the palindrome.

The idea is to concatenate word1 and word2 into one string and then apply a dynamic programming approach to find the longest palindrome subsequence. The dynamic programming (DP) solution uses a 2D table f where f[i][j] represents the length of the longest palindromic subsequence between index i and j in the concatenated string s. We initialize this table with zeros and fill it in a way that each cell f[i][j] takes into account the characters at s[i] and s[j]. If they match and belong to different parts (one from word1 and the other from word2), we can increase the length of a valid palindrome by 2. We iterate through the string in a manner such that we solve smaller subproblems first, which then can be used to construct solutions for larger subproblems.

While filling the DP table, we keep an eye out for the characters that come from the point where word1 ends and word2 begins. If s[i] and s[j] match and i belongs to word1 while j belongs to word2, it implies that we have a palindrome that uses characters from both words. This is when we update our answer (ans), which keeps track of the maximum length found so far.

In summary, the solution approach consists of the following steps:

1. Concatenate word1 and word2.
2. Initialize a DP table to store the length of the longest palindromic subsequence.
3. Update the table according to the matching characters in the string.
4. Keep track of the maximum length whenever characters in word1 and word2 contribute to the palindrome.
5. Return the maximum length recorded.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution is accomplished by applying dynamic programming (DP), which is an algorithmic technique used to solve problems by breaking them down into subproblems, solving each subproblem only once, and storing their solutions.

Here's a step-by-step explanation of the implementation:

1. Concatenating the Strings: The first step is to concatenate word1 and word2 into a single string s.

2. Initializing the DP Table: A 2D DP table f is initialized with dimensions n x n, where n is the length of the concatenated string s. Each cell f[i][j] in this table will eventually contain the length of the longest palindromic subsequence between indices i and j. For single characters, which can be considered as the shortest palindromes, the diagonal f[i][i] is filled with 1.

3. Bottom-Up DP Computation: The table is filled in a bottom-up manner. The loops are arranged to have i decrease from n-1 to 0 and j increase from i+1 to n-1. This ensures that before we try to compute f[i][j], the subproblems f[i+1][j-1], f[i+1][j], and f[i][j-1] are already computed.

4. Updating the DP Table:

   • Case When Characters Match: If s[i] matches s[j], it means that characters at these indices can potentially form or extend a palindrome. Therefore, we set f[i][j] = f[i+1][j-1] + 2. The term f[i+1][j-1] represents the length of the longest palindromic subsequence within the substring that excludes the current matching characters that are just added to the ends of the palindrome. The addition of 2 accounts for the two matching characters.

   • Checking the Words' Boundaries: If the matching characters are such that s[i] is in word1 and s[j] is in word2, it indicates the formation of a palindrome using parts of both word1 and word2. Thus, the current palindrome length f[i][j] is compared with the maximum length found so far, and if it's larger, ans is updated.

   • Case When Characters Don't Match: If s[i] does not match s[j], then the longest palindrome at f[i][j] is the maximum of the longest palindromic subsequences found by either including s[i] and excluding s[j] (f[i+1][j]), or excluding s[i] and including s[j] (f[i][j-1]). Essentially, this step omits one of the mismatching characters and looks at the best result achieved by the remaining string.

5. Returning the Result: The variable ans keeps track of the maximum length of a palindrome that can be constructed by using the aforementioned rules. After the loops are finished, ans contains the length of the longest palindrome possible, and that is what is returned by the function.

By employing this DP approach, we can efficiently compute the longest palindromic sequence that can be constructed from word1 and word2 by reusing solutions to subproblems and building up to the final solution.

Example Walkthrough

Let's say we have word1 = "abaxy" and word2 = "azby" and we want to find the length of the longest palindrome that can be formed by concatenating subsequences of these two words.

Following the steps of the solution:

1. Concatenating the Strings: By concatenating word1 and word2, we get s = "abaxyazby".

2. Initializing the DP Table: We create a 2D table f with dimensions 10 x 10, as s has a length of 10.

3. Bottom-Up DP Computation: We traverse the table by starting from the end of s and working backwards, ensuring that the substring problems are solved first.

4. Updating the DP Table:

   • As we iterate over the table, we check for every pair s[i] and s[j].
   • For i = 4 (the last character of word1) and j = 5 (the first character of word2):
     • We note that s[i] = y and s[j] = a do not match, so f[4][5] will be the max of f[5][5] and f[4][4], which are both 1, so f[4][5] = 1.
   • Continuing the process:
     • For i = 1 and j = 8, we see that s[i] = b and s[j] = b do match. Therefore, we set f[1][8] = f[2][7] + 2. Assume f[2][7] was already computed and is 1, so f[1][8] = 1 + 2 = 3. This is a valid palindrome consisting of the substrings "b" from word1 and "b" from word2.
   • We continue filling out the table in this manner, looking for matches and updating the length of potential palindromes accordingly.

5. Returning the Result: Assume after iterating through the entire table and considering all possible subsequence pairs, the maximum length recorded in our ans is 3 (from the palindrome "aba" or "bab"). This means the longest palindrome that can be formed by concatenating subsequences from word1 and word2 is of length 3.

In this example, the steps have given us a systematic approach to finding matches between two words and using these to find a palindrome by employing dynamic programming to avoid redundant calculations.

Solution Implementation

Time and Space Complexity

Time Complexity

The function longestPalindrome uses dynamic programming to solve the problem of finding the longest palindromic subsequence that can be formed using the concatenation of word1 and word2. The time complexity can be analyzed as follows:

• There is a nested loop structure where the outer loop runs for n times, where n is the length of the concatenated string s (that is, n = len(word1) + len(word2)). The inner loop also runs for n times, although not all iterations are full due to the starting point of the inner loop depending on the outer loop.
• In each iteration, operations are performed with constant time complexity, specifically: assignment, comparison, and arithmetic operations.

Therefore, the total time complexity is O(n^2), where n is the length of the concatenated string s.

Space Complexity

The space complexity of the function can be determined as follows:

• A 2D list f of size n * n is created to store the lengths of palindromic subsequences, where n is the length of the string s. Each cell of the list stores a single integer.
• Constant space is used for variables like i, j, ans, and for some temporary variables used during comparisons and assignments.

As the dominant space requirement is the 2D list f, the space complexity is O(n^2), where n is the length of the concatenated string s.

Learn more about how to find time and space complexity quickly using problem constraints.