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Problem solution · Python

Find Beautiful Indices in the Given Array II

Find Beautiful Indices in the Given Array 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
54 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 Find Beautiful Indices in the Given Array 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

  • 54 lines of Python from the credited upstream file 3008.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 codeFind Beautiful Indices in the Given Array II · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  # Same as 3006. Find Beautiful Indices in the Given Array I  def beautifulIndices(self, s: str, a: str, b: str, k: int) -> list[int]:    ans = []    indicesA = self._kmp(s, a)    indicesB = self._kmp(s, b)    indicesBIndex = 0  # indicesB' index     for i in indicesA:      # The constraint is: |j - i| <= k. So, -k <= j - i <= k. So, move      # `indicesBIndex` s.t. j - i >= -k, where j := indicesB[indicesBIndex].      while indicesBIndex < len(indicesB) and indicesB[indicesBIndex] - i < -k:        indicesBIndex += 1      if indicesBIndex < len(indicesB) and indicesB[indicesBIndex] - i <= k:        ans.append(i)     return ans   def _kmp(self, s: str, pattern: str) -> list[int]:    """Returns the starting indices of all occurrences of the pattern in `s`."""     def getLPS(pattern: str) -> list[int]:      """      Returns the lps array, where lps[i] is the length of the longest prefix of      pattern[0..i] which is also a suffix of this substring.      """      lps = [0] * len(pattern)      j = 0      for i in range(1, len(pattern)):        while j > 0 and pattern[j] != pattern[i]:          j = lps[j - 1]        if pattern[i] == pattern[j]:          lps[i] = j + 1          j += 1      return lps     lps = getLPS(pattern)    res = []    i = 0  # s' index    j = 0  # pattern's index    while i < len(s):      if s[i] == pattern[j]:        i += 1        j += 1        if j == len(pattern):          res.append(i - j)          j = lps[j - 1]      elif j != 0:  # Mismatch after j matches.        # Don't match lps[0..lps[j - 1]] since they will match anyway.        j = lps[j - 1]      else:        i += 1    return res

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