Problem Description

You are given a string s and an integer k.

In one operation, you can replace a 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.

Intuition

The task requires us to find the longest palindromic subsequence in a string s after applying at most k operations, where each operation is a character replacement to its neighboring letter in the alphabet with wrapping around.

The solution uses a dynamic programming approach to achieve this. We define a function dfs(i, j, k) that returns the length of the longest palindromic subsequence in the substring s[i..j] with at most k operations. The idea is to break down the problem for a given substring as follows:

1. Base Cases:

   • If the start index i exceeds the end index j (i > j), the longest palindromic subsequence is 0 because there are no characters left.
   • If the substring length is 1 (i == j), the longest palindromic subsequence is 1 as a single character is always a palindrome.

2. Recursive Cases:

   • We consider ignoring either the start (s[i]) or the end character (s[j]). This is done by calculating recursively for dfs(i + 1, j, k) or dfs(i, j - 1, k) respectively.
   • Alternatively, we can attempt to match s[i] and s[j] by performing necessary operations. We compute the difference t between the ASCII values of s[i] and s[j], considering the wrap around the z-a boundary. If t operations are feasible (t <= k), we can try forming a palindrome by considering dfs(i + 1, j - 1, k - t) + 2, where this new palindrome includes s[i] and s[j].

By considering these options, we compute the maximum between ignoring characters or forming a palindrome, and return the best value. Memoization is employed to store results of subproblems and avoid redundant calculations, thus optimizing the search process efficiently.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution to the problem employs a recursive approach enhanced by memoization to find the longest palindromic subsequence within a given string s, allowing up to k character modifications. Here’s a detailed breakdown of the implementation approach:

1. Recursive Function with Memoization:

   • We define a helper function dfs(i, j, k) where i and j are the indices defining the substring s[i..j] and k is the number of allowed operations left. This function calculates the longest palindromic subsequence within the given parameters.
   • The function is memoized using Python's @cache decorator, which helps store the results of previously computed states and improves efficiency by avoiding redundant calculations.

2. Base and Recursive Cases:

   • Base Case 1: If i > j, return 0 as there are no characters left to form a palindrome.
   • Base Case 2: If i == j, return 1 because a single character is inherently a palindrome.
   • Recursive Case: For each substring:
     • Compute the best result by either skipping a character from the start or the end (dfs(i + 1, j, k) or dfs(i, j - 1, k) respectively).
     • Calculate the difference d = abs(s[i] - s[j]) to determine the number of operations required to make s[i] equal to s[j]. Use wrap-around logic to account for the circular nature of the alphabet: t = min(d, 26 - d).
     • If t <= k, attempt to form a palindrome by matching s[i] and s[j] and calculate dfs(i + 1, j - 1, k - t) + 2.

3. Implementation Details:

   • Convert characters to their ASCII values using list(map(ord, s)) to facilitate the calculation of differences and modification requirements.
   • Initiate the recursive process with dfs(0, n - 1, k) where n is the length of s.
   • After computing the required result, clear the cache using dfs.cache_clear() to manage memory usage effectively.

This approach ensures that each potential subsequence is evaluated and the best palindrome is found under the constraint of k operations.

Example Walkthrough

Let's consider a simple example where the string s is "abc" and k is 1. We want to find the length of the longest palindromic subsequence after at most 1 operation.

1. Initial Setup:

   • We will use a recursive function, dfs(i, j, k), to calculate the longest palindromic subsequence between indices i and j, allowing up to k operations.
   • Initialize the process by calling dfs(0, 2, 1), where 0 and 2 are the starting and ending indices of the string.

2. Base Cases:

   • For any call where i > j, return 0 since there are no characters left to form a palindrome.
   • If i == j, return 1 as a single character is always a palindromic subsequence.

3. Recursive Cases:

   • Consider the substring "abc":
     • Call dfs(1, 2, 1) (ignoring 'a') and dfs(0, 1, 1) (ignoring 'c').
     • Compute the difference d between ASCII values of 'a' (97) and 'c' (99). The difference d = abs(97 - 99) = 2 (requires 2 operations to equalize, using wrap-around logic: t = min(2, 26 - 2) = 2).
     • Since t > k, we cannot match 'a' and 'c' with the current operation limit k = 1.

4. Evaluation of Subproblems:

   • Evaluate dfs(1, 2, 1) for the substring "bc":
     • Call dfs(2, 2, 1) and dfs(1, 1, 1) to consider cases ignoring either side.
     • Compute d = abs(98 - 99) = 1 (requires 1 operation to match 'b' and 'c', t = 1).
     • Perform the operation as t <= k and call dfs(2, 1, 0) + 2, resulting in a subsequence of length 2.

5. Combine and Conclude:

   • Backtrack to combine results according to maximum lengths obtained from all subproblems.
   • Finally, the length of the longest palindromic subsequence for "abc" with k=1 is 2 (e.g., forming "aba", "cbc" by replacing a single character).

By using a memoized recursive approach, the solution efficiently evaluates and returns the maximum possible length of a palindromic subsequence within the given constraints.

Solution Implementation

Time and Space Complexity

The given code uses a dynamic programming approach with memoization (cache) to solve the longest palindromic subsequence problem with an additional constraint.

Time Complexity

The function dfs(i, j, k) recursively explores all possible subsequences within indices i and j with a constraint involving k.

1. There are O(n^2) pairs of (i, j) since for each i, j can vary from i to n-1.
2. For each pair of indices (i, j), the function evaluates different values of k, and k can vary from 0 to the initial given k.

Combining these factors, the time complexity is O(n^2 * k).

Space Complexity

1. The space required to store intermediate results of dfs(i, j, k) in the cache is O(n^2 * k) because we store computations for each combination of (i, j, k).

2. The recursive stack requires O(n) space in the worst case due to the depth of recursion.

Thus, the space complexity is O(n^2 * k).

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