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Problem solution · Python

Check if an Original String Exists Given Two Encoded Strings

Check if an Original String Exists Given Two Encoded Strings: 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
52 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 Check if an Original String Exists Given Two Encoded Strings, 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

  • 52 lines of Python from the credited upstream file 2060.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 codeCheck if an Original String Exists Given Two Encoded Strings · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def possiblyEquals(self, s1: str, s2: str) -> bool:    def getNums(s: str) -> set[int]:      nums = {int(s)}      for i in range(1, len(s)):        nums |= {x + y for x in getNums(s[:i]) for y in getNums(s[i:])}      return nums     def getNextLetterIndex(s: str, i: int) -> int:      j = i      while j < len(s) and s[j].isdigit():        j += 1      return j     @functools.lru_cache(None)    def dp(i: int, j: int, paddingDiff: int) -> bool:      """      Returns True if s1[i..n) matches s2[j..n), accounting for the padding      difference. Here, `paddingDiff` represents the signed padding. A positive      `paddingDiff` indicates that s1 has an additional number of offset bytes      compared to s2.      """      if i == len(s1) and j == len(s2):        return paddingDiff == 0      # Add padding on s1.      if i < len(s1) and s1[i].isdigit():        nextLetterIndex = getNextLetterIndex(s1, i)        for num in getNums(s1[i:nextLetterIndex]):          if dp(nextLetterIndex, j, paddingDiff + num):            return True      # Add padding on s2.      elif j < len(s2) and s2[j].isdigit():        nextLetterIndex = getNextLetterIndex(s2, j)        for num in getNums(s2[j:nextLetterIndex]):          if dp(i, nextLetterIndex, paddingDiff - num):            return True      # `s1` has more padding, so j needs to catch up.      elif paddingDiff > 0:        if j < len(s2):          return dp(i, j + 1, paddingDiff - 1)      # `s2` has more padding, so i needs to catch up.      elif paddingDiff < 0:        if i < len(s1):          return dp(i + 1, j, paddingDiff + 1)      # There's no padding difference, so consume the next letter.      else:  # paddingDiff == 0        if i < len(s1) and j < len(s2) and s1[i] == s2[j]:          return dp(i + 1, j + 1, 0)      return False     return dp(0, 0, 0)

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