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Problem solution · Python

String Compression II

String Compression II: 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
37 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 String Compression II, 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

  • 37 lines of Python from the credited upstream file 1531.py.
  • The implementation visibly relies on hash lookup, 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 codeString Compression II · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def getLengthOfOptimalCompression(self, s: str, k: int) -> int:    def getLength(maxFreq: int) -> int:      """Returns the length to compress `maxFreq`."""      if maxFreq == 1:        return 1  # c      if maxFreq < 10:        return 2  # [1-9]c      if maxFreq < 100:        return 3  # [1-9][0-9]c      return 4    # [1-9][0-9][0-9]c     @functools.lru_cache(None)    def dp(i: int, k: int) -> int:      """Returns the length of optimal dp of s[i..n) with at most k deletion."""      if k < 0:        return math.inf      if i == len(s) or len(s) - i <= k:        return 0       ans = math.inf      maxFreq = 0  # the maximum frequency in s[i..j]      count = collections.Counter()       # Make letters in s[i..j] be the same.      # Keep the letter that has the maximum frequency in this range and remove      # the other letters.      for j in range(i, len(s)):        count[s[j]] += 1        maxFreq = max(maxFreq, count[s[j]])        ans = min(ans, getLength(maxFreq) +                  dp(j + 1, k - (j - i + 1 - maxFreq)))       return ans     return dp(0, k)

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