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Problem solution · Python

Strong Password Checker

Strong Password Checker: 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
48 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 Strong Password Checker, 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

  • 48 lines of Python from the credited upstream file 420.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 codeStrong Password Checker · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def strongPasswordChecker(self, password: str) -> int:    n = len(password)    missing = self._getMissing(password)    # the number of replacements to deal with 3 repeating characters    replaces = 0    # the number of sequences that can be substituted with 1 deletions,    # (3k)-seqs    oneSeq = 0    # the number of sequences that can be substituted with 2 deletions,    # (3k + 1)-seqs    twoSeq = 0     i = 2    while i < n:      if password[i] == password[i - 1] and password[i - 1] == password[i - 2]:        length = 2  # the length of the repeating password        while i < n and password[i] == password[i - 1]:          length += 1          i += 1        replaces += length // 3  # 'aaaaaaa' -> 'aaxaaxa'        if length % 3 == 0:          oneSeq += 1        if length % 3 == 1:          twoSeq += 1      else:        i += 1     if n < 6:      return max(6 - n, missing)    if n <= 20:      return max(replaces, missing)     deletes = n - 20    # Each replacement in (3k)-seqs can be substituted with 1 deletions.    replaces -= min(oneSeq, deletes)    # Each replacement in (3k + 1)-seqs can be substituted with 2 deletions.    replaces -= min(max(deletes - oneSeq, 0), twoSeq * 2) // 2    # Each replacement in other seqs can be substituted with 3 deletions.    replaces -= max(deletes - oneSeq - twoSeq * 2, 0) // 3    return deletes + max(replaces, missing)   def _getMissing(self, password: str) -> int:    return (3            - any(c.isupper() for c in password)            - any(c.islower() for c in password)            - any(c.isdigit() for c in password))

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