CPMagical CP Library

Problem solution · Python

Total Characters in String After Transformations II

Total Characters in String After Transformations II: 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
49 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 Total Characters in String After Transformations II, 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

  • 49 lines of Python from the credited upstream file 3337.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 codeTotal Characters in String After Transformations II · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  # Similar to 3335. Total Characters in String After Transformations I  def lengthAfterTransformations(self, s: str, t: int, nums: list[int]) -> int:    MOD = 1_000_000_007     def matrixMult(A: list[list[int]], B: list[list[int]]) -> list[list[int]]:      """Returns A * B."""      sz = len(A)      C = [[0] * sz for _ in range(sz)]      for i in range(sz):        for j in range(sz):          for k in range(sz):            C[i][j] += A[i][k] * B[k][j]            C[i][j] %= MOD      return C     def matrixPow(M: list[list[int]], n: int) -> list[list[int]]:      """Returns M^n."""      if n == 0:        return [[1 if i == j else 0  # identity matrix                for j in range(len(M))]                for i in range(len(M))]      if n % 2 == 1:        return matrixMult(M, matrixPow(M, n - 1))      return matrixPow(matrixMult(M, M), n // 2)     # T[i][j] := the number of ways to transform ('a' + i) to ('a' + j)    T = self._getTransformationMatrix(nums)    poweredT = matrixPow(T, t)    count = [0] * 26    lengths = [0] * 26     for c in s:      count[ord(c) - ord('a')] += 1     for i in range(26):      for j in range(26):        lengths[j] += count[i] * poweredT[i][j]        lengths[j] %= MOD     return sum(lengths) % MOD   def _getTransformationMatrix(self, nums: list[int]) -> list[list[int]]:    T = [[0] * 26 for _ in range(26)]    for i, steps in enumerate(nums):      for step in range(1, steps + 1):        T[i][(i + step) % 26] += 1    return T

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗