Introduction

In this blog post, we will explore how to solve the LeetCode problem "3472" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a string s and an integer k. In one operation, you can replace the character at any position with the next or previous letter in the alphabet (wrapping around so that 'a' is after 'z'). For example, replacing 'a' with the next letter results in 'b', and replacing 'a' with the previous letter results in 'z'. Similarly, replacing 'z' with the next letter results in 'a', and replacing 'z' with the previous letter results in 'y'. Return the length of the longest palindromic subsequence of s that can be obtained after performing at most k operations. Example 1: Input: s = "abced", k = 2 Output: 3 Explanation: Replace s[1] with the next letter, and s becomes "acced". Replace s[4] with the previous letter, and s becomes "accec". The subsequence "ccc" forms a palindrome of length 3, which is the maximum. Example 2: Input: s = "aaazzz", k = 4 Output: 6 Explanation: Replace s[0] with the previous letter, and s becomes "zaazzz". Replace s[4] with the next letter, and s becomes "zaazaz". Replace s[3] with the next letter, and s becomes "zaaaaz". The entire string forms a palindrome of length 6. Constraints: 1 <= s.length <= 200 1 <= k <= 200 s consists of only lowercase English letters.

Explanation

Here's the breakdown:

• Dynamic Programming: The core idea is to use dynamic programming to build up solutions for smaller subproblems. dp[i][j] represents the length of the longest palindromic subsequence of s[i:j+1] (inclusive) that can be obtained with at most k operations.

• Cost Calculation: Efficiently determine the cost (number of operations) required to make two characters equal by calculating the minimum distance between them in the alphabet.

• Palindrome Check and DP Update: If the cost to make s[i] equal to s[j] is within our allowed k, we can extend a smaller palindromic subsequence s[i+1:j-1] by 2. Otherwise, we consider the best palindromic subsequence from s[i+1:j] or s[i:j-1].

• Time & Space Complexity: O(n^2) time, O(n^2) space, where n is the length of the string s.

Code

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

    def cost(c1, c2):
        diff = abs(ord(c1) - ord(c2))
        return min(diff, 26 - diff)

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

    for length in range(2, n + 1):
        for i in range(n - length + 1):
            j = i + length - 1
            ops = cost(s[i], s[j])
            if ops <= k:
                if length == 2:
                    dp[i][j] = 2
                else:
                    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][n - 1]