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Problem solution · Python

Longest Palindromic Subsequence After at Most K Operations

Longest Palindromic Subsequence After at Most K Operations: a Python solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
24 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For Longest Palindromic Subsequence After at Most K Operations, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 24 lines of Python from the credited upstream file 3472.py.
  • The implementation visibly relies on cached states.
  • No explicit loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeLongest Palindromic Subsequence After at Most K Operations · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  # Similar to 516. Longest Palindromic Subsequence  def longestPalindromicSubsequence(self, s: str, k: int) -> int:    @functools.lru_cache(None)    def dp(i: int, j: int, op: int) -> int:      """Returns the length of LPS(s[i..j]) with at most `op` operations."""      if i > j:        return 0      if i == j:        return 1      if s[i] == s[j]:        return 2 + dp(i + 1, j - 1, op)      res = max(dp(i + 1, j, op), dp(i, j - 1, op))      cost = self._getCost(s[i], s[j])      if cost <= op:        res = max(res, 2 + dp(i + 1, j - 1, op - cost))      return res     return dp(0, len(s) - 1, k)   def _getCost(self, a: str, b: str) -> int:    dist = abs(ord(a) - ord(b))    return min(dist, 26 - dist)

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