Longest Palindromic Subsequence Leetcode Problem 516 [Python Solution]

The Longest Palindromic Subsequence problem on LeetCode (Problem 516) challenges us to find the length of the longest palindromic subsequence within a given string s.

But what exactly is a “palindromic subsequence”?

A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

In this context, we are looking for palindromic subsequences, which means the characters should form a palindrome when arranged in the same order.

Let’s take a look at an example to illustrate the problem:

Example 1:

Input: s = “bbbab”

Output: 4

Explanation: One possible longest palindromic subsequence is “bbbb”.

In this example, the string “bbbab” contains multiple subsequences, but the longest palindromic subsequence is “bbbb,” with a length of 4.

Example 2:

Input: s = “cbbd”

Output: 2

Explanation: One possible longest palindromic subsequence is “bb”.

Constraints:

• 1 <= s.length <= 1000
• s consists only of lowercase English letters.

The problem is clear, but how do we approach solving it?

There are multiple ways to tackle this problem, and we’ll explore two different solutions.

Understanding the Constraints

Before diving into the solutions, let’s take a moment to understand the constraints of this problem.

The input string s can be as long as 1000 characters and consists only of lowercase English letters.

This means we have a relatively large input space to consider, and our solution should be efficient enough to handle such input sizes within a reasonable time frame.

Brute Force Approach

One way to solve this problem is through a brute force approach.

We can explore all possible subsequences of the input string s and check if each subsequence is a palindrome.

The length of the longest palindromic subsequence will be the maximum length we encounter while iterating through these subsequences.

Let’s break down the steps of this brute force approach:

1. Initialize a variable max_length to store the length of the longest palindromic subsequence.

Set it to 0 initially.

2. Generate all possible subsequences of the input string s.

This can be done using recursion or bit manipulation.

3. For each generated subsequence, check if it’s a palindrome.

If it is, update the max_length if the length of the subsequence is greater than the current max_length.

4. After checking all subsequences, max_length will contain the length of the longest palindromic subsequence.

This brute force approach will give us the correct answer, but it’s highly inefficient, especially for large input strings.

The time complexity of this approach is exponential, and it’s not suitable for the constraints mentioned.

Efficient Approach with Dynamic Programming

To solve this problem efficiently, we’ll use dynamic programming, specifically a bottom-up approach, to avoid recalculating the same subproblems.

This approach will have a time complexity of O(n^2), where n is the length of the input string s.

The idea behind the dynamic programming solution is to maintain a 2D array dp, where dp[i][j] represents the length of the longest palindromic subsequence in the substring s[i:j+1].

We’ll fill in this dp table iteratively, building on previously computed values.

Here’s how the efficient dynamic programming solution works:

1. Initialize a 2D array dp of size (n x n), where n is the length of the input string s.

Initialize all elements of dp to 0.

2. Iterate through the input string s from right to left, and for each character at index i, iterate from that character to the end of the string.

This double iteration allows us to consider all possible substrings.

3. For each pair of indices (i, j), check if the characters s[i] and s[j] are equal.

If they are, it means we can potentially extend a palindromic subsequence.

We have two cases to consider:
a.

If i is equal to j, it means we have a single character, and the length of the palindromic subsequence is 1 (i.e., dp[i][j] = 1).
b.

If i is not equal to j, it means we can extend a palindromic subsequence if we add these two characters.

In this case, we increase the length by 2, indicating that we found two matching characters (dp[i][j] = dp[i+1][j-1] + 2).

4. If the characters s[i] and s[j] are not equal, we have to decide whether to move the pointer i to the right or j to the left to explore both possibilities.

We choose the maximum of these two values:
a. dp[i][j] = max(dp[i+1][j], dp[i][j-1])

5. Keep track of the maximum length found during the iteration, as this represents the length of the longest palindromic subsequence.
6. After completing the iteration, dp[0][n-1] will contain the length of the longest palindromic subsequence in the entire string s.

The dynamic programming solution efficiently computes the answer by avoiding redundant calculations and exploring all possible palindromic subsequences within the given constraints.

Let’s take a look at the Python code for this dynamic programming solution:

class Solution:
    def longestPalindromeSubseq(self, s: str) -> int:
        n = len(s)
        dp = [[0] * n for _ in range(n)]

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

        return dp[0][n - 1]

This code efficiently calculates the length of the longest palindromic subsequence in the input string s.

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Reasoning Behind Our Efficient Approach

The dynamic programming solution is based on the idea of breaking down the problem into smaller subproblems and reusing the results of those subproblems to build the final solution.

By maintaining a 2D array dp to store the lengths of palindromic subsequences in various substrings, we avoid redundant calculations and make the solution more efficient.

The key insight is that for a substring s[i:j+1] (where i and j are indices within the string), the length of the longest palindromic subsequence can be determined by looking at the characters at i and j and the solution for the subproblem

s[i+1:j] and s[i:j-1].

By considering these overlapping subproblems, we build a dynamic programming table that gradually fills in the lengths of palindromic subsequences for various substrings.

The final answer is obtained in dp[0][n-1], where n is the length of the input string.

This approach is efficient and guarantees a time complexity of O(n^2), making it suitable for solving the problem within the given constraints.

Time Complexity: The time complexity of the efficient dynamic programming solution is O(n^2), where n is the length of the input string s.

We perform a double iteration through the string to fill in the dp table, and each iteration takes O(n) time.

Therefore, the overall time complexity is O(n^2).

Space Complexity: The space complexity is O(n^2) as well.

We use a 2D array dp of size (n x n) to store the lengths of palindromic subsequences in various substrings.

This space is required to cache the results of subproblems and build the dynamic programming table.

By using dynamic programming, we efficiently solve the Longest Palindromic Subsequence problem, ensuring that we can handle input strings of reasonable length within the given constraints.

In summary, we’ve explored two approaches to solving the Longest Palindromic Subsequence problem on LeetCode.

While the brute force approach is conceptually simple, it’s not suitable for large input sizes.

The efficient dynamic programming approach, on the other hand, is designed to handle the constraints efficiently and provides a correct solution.

Remember that dynamic programming is a powerful technique for solving a wide range of algorithmic problems.

Understanding its principles and applications can be valuable for your coding interviews and competitive programming challenges.

If you have any questions, suggestions, or comments, please feel free to share them in the comments section.

We encourage you to engage with the content, ask questions, and provide feedback.

Additionally, don’t forget to check out the LeetCode problem link to practice and further refine your coding skills.

LeetCode Problem 516 – Longest Palindromic Subsequence

Thank you for reading, and happy coding!