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Problem solution · Python

Subsequence With the Minimum Score

Subsequence With the Minimum Score: 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
47 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 Subsequence With the Minimum Score, 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

  • 47 lines of Python from the credited upstream file 2565.py.
  • The implementation keeps its working state in language-native values and containers.
  • 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 codeSubsequence With the Minimum Score · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def minimumScore(self, s: str, t: str) -> int:    # leftmost[j] := the minimum index i s.t. t[0..j] is a subsequence of s[0..i].    #          -1 := impossible    leftmost = [-1] * len(t)    # rightmost[j] := the maximum index i s.t. t[j:] is a subsequence of s[i..n).    #           -1 := impossible    rightmost = [-1] * len(t)     j = 0  # t's index    for i in range(len(s)):      if s[i] == t[j]:        leftmost[j] = i        j += 1        if j == len(t):          break     j = len(t) - 1  # t's index    for i in reversed(range(len(s))):      if s[i] == t[j]:        rightmost[j] = i        j -= 1        if j == -1:          break     # The worst case is to delete t[0:j] since t[j:] is a subsequence of s. (deduced    # from the above loop).    ans = j + 1     j = 0    for i in range(len(t)):      # It's impossible that t[0..i] is a subsequence of s. So, stop the loop since      # no need to consider any larger i.      if leftmost[i] == -1:        break      # While t[0..i] + t[j:] is not a subsequence of s, increase j.      while j < len(t) and leftmost[i] >= rightmost[j]:        j += 1      # Now, leftmost[i] < rightmost[j], so t[0..i] + t[j:n] is a subsequence of s.      # If i == j that means t is a subsequence of s, so just return 0.      if i == j:        return 0      # Delete t[i + 1..j - 1] and that's a total of j - i - 1 letters.      ans = min(ans, j - i - 1)     return ans

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