Problem Description

You are given a string s and an integer k.

You can perform operations on the string where each operation allows you to replace any character with the next or previous letter in the alphabet. The alphabet wraps around, meaning:

• 'a' becomes 'b' when moved forward or 'z' when moved backward
• 'z' becomes 'a' when moved forward or 'y' when moved backward

Your goal is to find the length of the longest palindromic subsequence that can be obtained from s after performing at most k operations.

A subsequence is a sequence that can be derived from the string by deleting some or no characters without changing the order of the remaining characters. A palindromic subsequence reads the same forwards and backwards.

For example:

• If you have string "abc" with k = 2, you could change 'c' to 'a' (2 operations: c→b→a), making the string "aba". The longest palindromic subsequence would be "aba" with length 3.

The key insight is that when trying to match two characters s[i] and s[j] to form part of a palindrome, the minimum number of operations needed is the shorter distance between them on the circular alphabet. For instance, to change 'a' to 'z', you can go either forward 25 steps or backward 1 step, so the cost is min(25, 1) = 1.

The solution uses dynamic programming with memoization, where dfs(i, j, k) represents the maximum palindrome length achievable in substring s[i..j] with at most k operations remaining. At each step, you can either:

1. Skip character at position i or j
2. Try to match characters at positions i and j if the cost is within budget

Intuition

The problem asks for the longest palindromic subsequence, which immediately suggests a dynamic programming approach similar to the classic longest palindromic subsequence problem. However, there's a twist - we can modify characters with a limited budget of k operations.

The key insight is recognizing that in a palindrome, characters at symmetric positions must match. When building a palindrome, we compare characters from both ends of our current range. If they already match, great! If not, we need to decide whether it's worth spending operations to make them match.

For any two characters, the cost to make them identical is the minimum distance between them on the circular alphabet. Since the alphabet wraps around, we can go either clockwise or counterclockwise. For example, to change 'a' to 'd', we can go forward 3 steps (a→b→c→d) or backward 23 steps - so we choose the minimum: min(3, 23) = 3.

This leads us to a recursive approach where for each substring s[i..j], we have three choices:

1. Exclude the left character: Find the longest palindrome in s[i+1..j] with the same budget k
2. Exclude the right character: Find the longest palindrome in s[i..j-1] with the same budget k
3. Try to match both ends: If we can afford to make s[i] and s[j] the same (cost ≤ k), we gain 2 characters for our palindrome and recursively solve for s[i+1..j-1] with the remaining budget

The maximum of these three options gives us the longest palindrome for the current substring with the given budget. We use memoization to avoid recalculating the same subproblems, caching results based on the triplet (i, j, k).

The base cases are straightforward: an empty range returns 0, and a single character is always a palindrome of length 1.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution implements a memoized recursive search (top-down dynamic programming) to find the longest palindromic subsequence.

Implementation Details:

1. Character Conversion: First, we convert the string to a list of ASCII values using list(map(ord, s)). This makes calculating the distance between characters easier since we can use simple arithmetic operations.

2. Main Recursive Function dfs(i, j, k):

   • Parameters:
     • i: left boundary of the current substring
     • j: right boundary of the current substring
     • k: remaining operations budget
   • Returns: length of the longest palindromic subsequence in s[i..j] using at most k operations

3. Base Cases:

   • If i > j: The range is invalid, return 0
   • If i == j: Single character is always a palindrome, return 1

4. Recursive Cases:

   • Option 1: Skip the left character - dfs(i + 1, j, k)
   • Option 2: Skip the right character - dfs(i, j - 1, k)
   • Option 3: Try to match both ends if affordable:
     • Calculate the distance: d = abs(s[i] - s[j])
     • Find minimum cost considering circular alphabet: t = min(d, 26 - d)
     • If t <= k, we can afford to match them: dfs(i + 1, j - 1, k - t) + 2
   • Return the maximum of all valid options

5. Distance Calculation: The cost to change one character to another is the minimum of:

   • Direct distance: abs(s[i] - s[j])
   • Wraparound distance: 26 - abs(s[i] - s[j])

   For example, changing 'a' (97) to 'z' (122) costs min(25, 26-25) = min(25, 1) = 1

6. Memoization: The @cache decorator automatically memoizes the function results, preventing redundant calculations for the same (i, j, k) triplet.

7. Final Answer: Call dfs(0, n - 1, k) to solve for the entire string with the full budget.

The time complexity is O(n² × k) where n is the length of the string, as we have n² possible substrings and k different budget states. The space complexity is also O(n² × k) for the memoization cache.

Example Walkthrough

Let's walk through the solution with string s = "abcd" and k = 3.

Initial Setup:

• Convert string to ASCII values: [97, 98, 99, 100] (representing 'a', 'b', 'c', 'd')
• Start with dfs(0, 3, 3) - checking the entire string with budget 3

Step 1: dfs(0, 3, 3) - Can we form a palindrome from "abcd"?

• Option 1: Skip 'a' → dfs(1, 3, 3)
• Option 2: Skip 'd' → dfs(0, 2, 3)
• Option 3: Match 'a' and 'd'
  • Distance = |97 - 100| = 3
  • Circular distance = min(3, 26-3) = 3
  • Cost is 3, exactly our budget!
  • If we match them: 2 + dfs(1, 2, 0)

Step 2: Explore dfs(1, 2, 0) - "bc" with budget 0

• Option 1: Skip 'b' → dfs(2, 2, 0) = 1 (just 'c')
• Option 2: Skip 'c' → dfs(1, 1, 0) = 1 (just 'b')
• Option 3: Match 'b' and 'c'
  • Distance = |98 - 99| = 1
  • Cost is 1, but budget is 0, so we can't match
• Result: max(1, 1) = 1

Step 3: Back to dfs(0, 3, 3) Option 3:

• Matching 'a' and 'd' gives us: 2 + 1 = 3

Step 4: Explore dfs(1, 3, 3) - "bcd" with budget 3

• Option 1: Skip 'b' → dfs(2, 3, 3)
• Option 2: Skip 'd' → dfs(1, 2, 3)
• Option 3: Match 'b' and 'd'
  • Distance = |98 - 100| = 2
  • Cost is 2, we can afford it
  • Result: 2 + dfs(2, 2, 1) = 2 + 1 = 3

Step 5: Explore dfs(0, 2, 3) - "abc" with budget 3

• Option 1: Skip 'a' → dfs(1, 2, 3)
• Option 2: Skip 'c' → dfs(0, 1, 3)
• Option 3: Match 'a' and 'c'
  • Distance = |97 - 99| = 2
  • Cost is 2, we can afford it
  • Result: 2 + dfs(1, 1, 1) = 2 + 1 = 3

Final Result: All three main options give us a palindrome of length 3:

• Skipping 'a': "bcd" → can form "bdb" (length 3)
• Skipping 'd': "abc" → can form "aba" (length 3)
• Matching 'a' and 'd': forms "ada" with middle character (length 3)

The maximum is 3, which is our answer.

The key insight demonstrated here is that we explore all possibilities - either skipping characters to find natural palindromes in subsequences, or using our operation budget to force characters to match when beneficial.

Solution Implementation

Time and Space Complexity

The time complexity of this code is O(n² × k), where n is the length of the string s and k is the parameter representing the maximum allowed transformation cost.

This complexity arises from the recursive function dfs(i, j, k) with memoization:

• The function has three parameters: i ranges from 0 to n-1, j ranges from 0 to n-1, and k ranges from 0 to the initial value of k
• Due to the constraint i <= j in the problem logic, there are approximately n²/2 valid (i, j) pairs
• For each valid (i, j) pair, we can have up to k+1 different values for the third parameter
• Each state is computed only once due to the @cache decorator (memoization)
• Each state computation takes O(1) time for the operations inside the function

Therefore, the total number of unique states is O(n² × k), and each state is computed in O(1) time, giving us a time complexity of O(n² × k).

The space complexity is also O(n² × k) due to:

• The memoization cache stores all computed states, which is O(n² × k) entries
• The recursion stack depth is at most O(n) in the worst case (when we recurse from (0, n-1) down to base cases)
• Since O(n² × k) > O(n), the dominant space complexity is O(n² × k)

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

Common Pitfalls

1. Incorrect Distance Calculation for Circular Alphabet

One of the most common mistakes is incorrectly calculating the minimum distance between two characters in a circular alphabet. Many developers forget to consider the wraparound nature of the alphabet.

Incorrect Implementation:

# Wrong: Only considers forward distance
distance = abs(ord(s[i]) - ord(s[j]))

Why it's wrong: This only calculates the direct distance. For example, changing 'a' to 'z' would cost 25, but going backward (a→z) only costs 1.

Correct Implementation:

# Correct: Considers both forward and backward distances
char_diff = abs(ord(s[i]) - ord(s[j]))
distance = min(char_diff, 26 - char_diff)

2. Confusing Subsequence with Substring

Another pitfall is treating this as a substring problem rather than a subsequence problem.

Wrong Approach:

# Trying to find contiguous palindromes only
for length in range(n, 0, -1):
    for start in range(n - length + 1):
        if is_palindrome_possible(s[start:start+length], k):
            return length

Correct Approach: The solution correctly handles subsequences by having three choices at each step:

• Skip left character
• Skip right character
• Match both ends (if affordable)

3. Off-by-One Errors in Recursion

Be careful with index boundaries when recursing:

Common Mistake:

# Wrong: Missing the +2 when matching both ends
if circular_distance <= remaining_cost:
    match_both = find_longest_palindrome(left + 1, right - 1, remaining_cost - circular_distance)
    # Forgot to add 2 for the matched characters!

Correct:

if circular_distance <= remaining_cost:
    match_both = find_longest_palindrome(left + 1, right - 1, remaining_cost - circular_distance) + 2

4. Not Handling Edge Cases Properly

Missing Base Cases:

def dfs(i, j, k):
    # Wrong: Missing base cases
    result = max(dfs(i+1, j, k), dfs(i, j-1, k))
    # This will cause infinite recursion!

Correct Base Cases:

def dfs(i, j, k):
    if i > j:
        return 0  # Invalid range
    if i == j:
        return 1  # Single character
    # Then handle recursive cases...

5. Memory Limit Issues with Large Inputs

For very large inputs, the memoization cache can consume significant memory. The solution correctly addresses this by clearing the cache after getting the result:

answer = find_longest_palindrome(0, string_length - 1, k)
find_longest_palindrome.cache_clear()  # Important for memory management
return answer

Without this, repeated test cases could cause memory issues in competitive programming platforms.