Maximize Palindrome Length From Subsequences

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StringsDynamic Programming

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the input strings `word1` and `word2` contain any characters outside of lowercase English letters?
Are `word1` and `word2` guaranteed to contain at least one character, or can they be empty strings?
If no palindrome can be formed by combining subsequences from `word1` and `word2`, what value should I return?
Are we looking for the longest possible palindrome, or are there any constraints on *how* the palindrome is formed from the subsequences (e.g. must take the largest subsequence from each string)?
If multiple palindromes of the maximum length can be formed, is any one acceptable?

Brute Force Solution

Approach

The brute force approach to finding the longest palindrome involves checking absolutely every combination of characters. We essentially create every possible subsequence from the given strings and see if we can combine any of them into a palindrome. We then compare the lengths of all palindromes we created and select the longest one.

Here's how the algorithm would work step-by-step:

First, consider all possible subsequences that you can make from the first string. A subsequence is any combination of letters from that string, in the same order they appeared, but you don't have to use all the letters.
Do the same thing for the second string, making every possible subsequence.
Now, for every subsequence from the first string, try combining it with every subsequence from the second string. Concatenate each pair of subsequences.
For each combination, check if the resulting combined string is a palindrome. A palindrome reads the same forwards and backward.
Keep track of the length of every palindrome you find.
After you've tried all the combinations of subsequences, find the largest palindrome length that you saved. That's your answer.

Code Implementation

def maximize_palindrome_length_from_subsequences(word1, word2):
    def generate_subsequences(word):
        subsequences = ['']
        for char in word:
            new_subsequences = [subsequence + char for subsequence in subsequences]
            subsequences.extend(new_subsequences)
        return subsequences

    def is_palindrome(word):
        return word == word[::-1]

    max_palindrome_length = 0

    subsequences_word1 = generate_subsequences(word1)
    subsequences_word2 = generate_subsequences(word2)

    # Iterate through all subsequence combinations to find palindromes
    for subsequence_one in subsequences_word1:

        for subsequence_two in subsequences_word2:

            combined_word = subsequence_one + subsequence_two

            # Check if the combined subsequence is a palindrome
            if is_palindrome(combined_word):

                max_palindrome_length = max(max_palindrome_length, len(combined_word))

    return max_palindrome_length

Big(O) Analysis

Time Complexity

O(2^(2n)) – Generating all subsequences from the first string of length n takes O(2^n) time because each character can either be included or excluded. Similarly, generating all subsequences from the second string also takes O(2^n) time. Combining each subsequence from the first string with each subsequence from the second string leads to O(2^n * 2^n) = O(2^(2n)) concatenations. Checking if each concatenated string is a palindrome takes O(n) time in the worst case, however since the number of combinations dominates, the overall time complexity is still driven by the generation and combination of subsequences, thus the O(2^(2n)) term is what remains.

Space Complexity

O(2^(N+M)) – The brute force approach generates all possible subsequences from both strings. If the first string has length N and the second has length M, then the number of subsequences generated from the first string is 2^N and the number from the second string is 2^M. Combining each subsequence from the first string with each from the second requires storing all 2^N * 2^M combinations, meaning storing all possible concatenations. The space used to store these concatenated strings will be proportional to 2^N * 2^M, which can be simplified to O(2^(N+M)).

Optimal Solution

Approach

The goal is to find the longest possible palindrome by combining pieces from two given strings. We'll achieve this by looking for matching characters from the ends of the strings and working our way inwards, efficiently building the palindrome.

Here's how the algorithm would work step-by-step:

Start by checking if the last character of the first string matches the first character of the second string.
If they match, include both characters in our potential palindrome and then shrink both strings by removing the matched characters from their ends.
If they don't match, consider two options: remove the last character of the first string, or remove the first character of the second string. Determine which gives us the better chance to find more matches later on.
Repeat steps 2 and 3, always trying to match the ends of the remaining parts of the two strings, and making smart choices about which characters to discard when there isn't a match.
Keep going until either string is empty.
If both strings are empty, we've found the longest possible palindrome length. If only one string remains, try to find the longest palindromic subsequence of the single string remaining and add its length to the length found so far.
The total length of the palindrome constructed is the final answer.

Code Implementation

def longestPalindrome(word1, word2):
    combined_word = word1 + word2
    length_of_combined_word = len(combined_word)

    def longestPalindromicSubsequence(word):
        length_of_word = len(word)
        dp = [[0] * length_of_word for _ in range(length_of_word)]

        for i in range(length_of_word):
            dp[i][i] = 1

        for subsequence_length in range(2, length_of_word + 1):
            for i in range(length_of_word - subsequence_length + 1):
                j = i + subsequence_length - 1
                if word[i] == word[j]:
                    dp[i][j] = dp[i + 1][j - 1] + 2
                else:
                    dp[i][j] = max(dp[i + 1][j], dp[i][j - 1])

        return dp[0][length_of_word - 1]

    longest_palindrome_length = 0
    left_index = 0
    right_index = len(word1) - 1

    while left_index < len(word1) and right_index >= 0:
        if word1[left_index] == word2[len(word2) - 1 - right_index]:
            # Matching chars extend palindrome.
            longest_palindrome_length += 2
            left_index += 1
            right_index -= 1
        else:
            break

    remaining_word1 = word1[left_index:]
    remaining_word2 = word2[:len(word2) - 1 - right_index]

    # Combine remaining strings to find LPS
    remaining_combined_word = remaining_word1 + remaining_word2
    palindrome_length_of_remaining_word = longestPalindromicSubsequence(remaining_combined_word)

    # Return total palindrome length.
    return longest_palindrome_length + palindrome_length_of_remaining_word

Big(O) Analysis

Time Complexity

O(n²) – The described approach uses a recursive strategy (or dynamic programming equivalent) to compare subsequences of two strings. The core logic involves comparing the ends of the strings and, in case of a mismatch, exploring two possibilities: removing a character from the first string or removing a character from the second string. In the worst case, the algorithm explores all possible pairs of subsequences derived from the input strings, leading to a recursive tree where each node can branch into two subproblems. If n is the combined length of the two strings, the maximum depth of this tree is n, and for each node in this tree we have to compare characters to detect for palindrome, hence the algorithm has O(n²) time complexity because, in the worst case, it boils down to checking all possible pairs of substrings.

Space Complexity

O(N^2) – The algorithm described uses recursion. In the worst-case scenario, where characters never match, the recursive calls branch out, exploring different possibilities of removing characters from either string. The maximum depth of the recursion can be proportional to the sum of the lengths of the two input strings. Each level of the recursion stores the function call stack, along with input strings (or substrings in this case). Furthermore, step 6 mentions finding the longest palindromic subsequence of a single remaining string, and a dynamic programming approach to solving this typically requires O(N^2) space where N is length of the string. Therefore, the space complexity is O(N^2).

Edge Cases

Case	How to Handle
word1 or word2 is null or empty	Return 0 since an empty string cannot contribute to a palindrome.
word1 and word2 have no characters in common	Return 0 as no palindrome can be formed from disjoint subsequences.
word1 and word2 are identical	Consider the entire combined string, if it is a palindrome just return its full length.
One word is a palindrome itself	Iterate over all possible starting and ending points in the other string to find if the palindrome can be maximized.
Very long words exceeding reasonable memory limits	Be mindful of the space complexity of dynamic programming solutions to avoid memory exhaustion; consider optimizations such as rolling arrays.
The optimal solution only contains letters from one word	The algorithm must consider subsequences formed only from characters present in one or the other of the two initial words.
Both words contain the same repeating character (e.g. 'aaaa' and 'bbbb')	Iterate through one word to find the first available character in the other word that makes the longest subsequence.
Integer overflow when calculating length	Use appropriate data types (e.g., long) to store lengths of subsequences to prevent potential overflows.

word1 or word2 is null or empty

How to Handle:

Return 0 since an empty string cannot contribute to a palindrome.

word1 and word2 have no characters in common

How to Handle:

Return 0 as no palindrome can be formed from disjoint subsequences.

word1 and word2 are identical

How to Handle:

Consider the entire combined string, if it is a palindrome just return its full length.

One word is a palindrome itself

How to Handle:

Iterate over all possible starting and ending points in the other string to find if the palindrome can be maximized.

Very long words exceeding reasonable memory limits

How to Handle:

Be mindful of the space complexity of dynamic programming solutions to avoid memory exhaustion; consider optimizations such as rolling arrays.

The optimal solution only contains letters from one word

How to Handle:

The algorithm must consider subsequences formed only from characters present in one or the other of the two initial words.

Both words contain the same repeating character (e.g. 'aaaa' and 'bbbb')

How to Handle:

Iterate through one word to find the first available character in the other word that makes the longest subsequence.

Integer overflow when calculating length

How to Handle:

Use appropriate data types (e.g., long) to store lengths of subsequences to prevent potential overflows.