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Problem solution · Python

Minimum Cost to Convert String II

Minimum Cost to Convert 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
59 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 Minimum Cost to Convert 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

  • 59 lines of Python from the credited upstream file 2977.py.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup, 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 codeMinimum Cost to Convert String II · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def minimumCost(      self,      source: str,      target: str,      original: list[str],      changed: list[str],      cost: list[int],  ) -> int:    subLengths = set(len(s) for s in original)    subToId = self._getSubToId(original, changed)    subCount = len(subToId)    # dist[u][v] := the minimum distance to change the substring with id u to    # the substring with id v    dist = [[math.inf for _ in range(subCount)] for _ in range(subCount)]    # dp[i] := the minimum cost to change the first i letters of `source` into    # `target`, leaving the suffix untouched    dp = [math.inf for _ in range(len(source) + 1)]     for a, b, c in zip(original, changed, cost):      u = subToId[a]      v = subToId[b]      dist[u][v] = min(dist[u][v], c)     for k in range(subCount):      for i in range(subCount):        if dist[i][k] < math.inf:          for j in range(subCount):            if dist[k][j] < math.inf:              dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])     dp[0] = 0     for i, (s, t) in enumerate(zip(source, target)):      if dp[i] == math.inf:        continue      if s == t:        dp[i + 1] = min(dp[i + 1], dp[i])      for subLength in subLengths:        if i + subLength > len(source):          continue        subSource = source[i:i + subLength]        subTarget = target[i:i + subLength]        if subSource not in subToId or subTarget not in subToId:          continue        u = subToId[subSource]        v = subToId[subTarget]        if dist[u][v] != math.inf:          dp[i + subLength] = min(dp[i + subLength], dp[i] + dist[u][v])     return -1 if dp[len(source)] == math.inf else dp[len(source)]   def _getSubToId(self, original: str, changed: str) -> dict[str, int]:    subToId = {}    for s in original + changed:      if s not in subToId:        subToId[s] = len(subToId)    return subToId

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