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Problem solution · Python

Find the Original Typed String II

Find the Original Typed String 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
41 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 Find the Original Typed String 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

  • 41 lines of Python from the credited upstream file 3333.py.
  • The implementation visibly relies on sequence storage, 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 codeFind the Original Typed String II · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def possibleStringCount(self, word: str, k: int) -> int:    MOD = 1_000_000_007    groups = self._getConsecutiveLetters(word)    totalCombinations = functools.reduce(lambda subtotal, group:                                         subtotal * group % MOD, groups)    if k <= len(groups):      return totalCombinations     # dp[j] := the number of ways to form strings of length j using groups[0..i]    dp = [0] * k    dp[0] = 1  # Base case: empty string     for i, group in enumerate(groups):      newDp = [0] * k      windowSum = 0      for j in range(i, k):        newDp[j] = (newDp[j] + windowSum) % MOD        windowSum = (windowSum + dp[j]) % MOD        if j >= group:          windowSum = (windowSum - dp[j - group] + MOD) % MOD      dp = newDp     return (totalCombinations - sum(dp)) % MOD   def _getConsecutiveLetters(self, word: str) -> list[int]:    """    Returns consecutive identical letters in the input string.    e.g. "aabbbc" -> [2, 3, 1].    """    groups = []    group = 1    for i in range(1, len(word)):      if word[i] == word[i - 1]:        group += 1      else:        groups.append(group)        group = 1    groups.append(group)    return groups

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