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Leetcode #2299: Strong Password Checker II

In this guide, we solve Leetcode #2299 Strong Password Checker II in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

A password is said to be strong if it satisfies all the following criteria: It has at least 8 characters. It contains at least one lowercase letter.

Quick Facts

  • Difficulty: Easy
  • Premium: No
  • Tags: String

Intuition

We need to scan characters while tracking positions or counts.

A simple state machine keeps the logic precise.

Approach

Iterate through the string once and update the state for each character.

Use a map or array if you need fast lookups.

Steps:

  • Iterate through characters.
  • Maintain necessary state.
  • Build or validate the output.

Example

Input: password = "IloveLe3tcode!" Output: true Explanation: The password meets all the requirements. Therefore, we return true.

Python Solution

class Solution: def strongPasswordCheckerII(self, password: str) -> bool: if len(password) < 8: return False mask = 0 for i, c in enumerate(password): if i and c == password[i - 1]: return False if c.islower(): mask |= 1 elif c.isupper(): mask |= 2 elif c.isdigit(): mask |= 4 else: mask |= 8 return mask == 15

Complexity

The time complexity is O(n)O(n)O(n), and the space complexity is O(1)O(1)O(1). The space complexity is O(1)O(1)O(1).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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