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Problem solution · Python

Lexicographically Smallest Generated String

Lexicographically Smallest Generated String: 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
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

Direct simulation

For Lexicographically Smallest Generated String, 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

  • 52 lines of Python from the credited upstream file 3474.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 codeLexicographically Smallest Generated String · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def generateString(self, str1: str, str2: str) -> str:    n = len(str1)    m = len(str2)    sz = n + m - 1    ans = [None] * sz    modifiable = [True] * sz     # 1. Handle all 'T' positions first.    for i, tf in enumerate(str1):      if tf == 'T':        for j, c in enumerate(str2):          pos = i + j          if ans[pos] and ans[pos] != c:            return ''          ans[pos] = c          modifiable[pos] = False     # 2. Fill all remaining positions with 'a'.    for i in range(sz):      if not ans[i]:        ans[i] = 'a'     # 3. Handle all 'F' positions.    for i in range(n):      if str1[i] == 'F' and self._match(ans, i, str2):        modifiablePos = self._lastModifiablePosition(i, m, modifiable)        if modifiablePos == -1:          return ''        ans[modifiablePos] = 'b'        modifiable[modifiablePos] = False     return ''.join(ans)   def _match(self, ans: list, i: int, s: str) -> bool:    """Returns True if the substring of ans starting at `i` matches `s`."""    for j, c in enumerate(s):      if ans[i + j] != c:        return False    return True   def _lastModifiablePosition(self, i: int, m: int, modifiable: list) -> int:    """    Finds the last modifiable position in the substring of ans starting at `i`.    """    modifiablePos = -1    for j in range(m):      pos = i + j      if modifiable[pos]:        modifiablePos = pos    return modifiablePos

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