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Problem solution · Python

Stamping The Sequence

Stamping The Sequence: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
46 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

Direct simulation

For Stamping The Sequence, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 46 lines of Python from the credited upstream file 936.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 codeStamping The Sequence · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def movesToStamp(self, stamp: str, target: str) -> list[int]:    def stampify(s: int) -> int:      """      Stamps target[i..i + |stamp|) and returns the number of newly stamped      characters.      e.g. stampify("abc", "ababc", 2) returns 3 because target becomes "ab***".      """      stampified = len(stamp)       for i, st in enumerate(stamp):        if target[s + i] == '*':  # It's already been stamped.          stampified -= 1        elif target[s + i] != st:  # We can't stamp on the index i.          return 0       for i in range(s, s + len(stamp)):        target[i] = '*'       return stampified     ans = []    target = list(target)    # stamped[i] := True if we already stamped target by stamping on index i    stamped = [False] * len(target)    stampedCount = 0  # Our goal is to make stampedCount = |target|.     while stampedCount < len(target):      isStamped = False      # Try to stamp target[i..i + |stamp|) for each index.      for i in range(len(target) - len(stamp) + 1):        if stamped[i]:          continue        stampified = stampify(i)        if stampified == 0:          continue        stampedCount += stampified        isStamped = True        stamped[i] = True        ans.append(i)      # After trying to stamp on each index, we can't find a valid stamp.      if not isStamped:        return []     return ans[::-1]

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