Longest Palindromic Subsequence (LPS)

Welcome to this in-depth guide on the Longest Palindromic Subsequence (LPS)! 🎉

In this lesson, we'll explore the Longest Palindromic Subsequence problem, a classic algorithm problem that asks for the longest palindromic sequence within a given string. This problem is crucial for understanding dynamic programming and string manipulation.

What is a Palindrome? 💡

A palindrome is a sequence that reads the same backward as forward. For example: "racecar", "madam", "level".

The Longest Palindromic Subsequence Problem 📝

Given a string, find the longest palindromic subsequence (not necessarily a palindrome itself). For instance, the longest palindromic subsequence for the string "banana" is "anana".

How to Solve the LPS Problem 🎯

The LPS problem can be solved using dynamic programming. We create an n x n matrix, where dp[i][j] represents the length of the longest palindromic subsequence ending at i and j.

Dynamic Programming Approach 📝

Initialize the matrix with all elements as 0.
Iterate from i=0 to n-1 for the first row, and from j=0 to n-1 for the first column.
For the rest of the matrix, calculate each cell using the following formula:

python

if (s[i] == s[j]):
    dp[i][j] = 1 + dp[i+1][j-1]
else:
    dp[i][j] = max(dp[i][j-1], dp[i+1][j])

The maximum element in the last row is the length of the longest palindromic subsequence.

Code Examples 💻

Here's a Python example for the LPS problem:

python

def longest_palindromic_subsequence(s):
    n = len(s)
    dp = [[0]*n for _ in range(n)]

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

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

    return dp[0][-1]

Let's dive deeper into the LPS problem, explore its applications, and even try to optimize the solution. Happy coding! 🚀