How do you solve Strong Password Checker II on LeetCode?

Strong Password Checker II (LeetCode #2299) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Checking all conditions in a single pass is more efficient than multiple passes.

What is the brute force solution for Strong Password Checker II?

The brute force approach for Strong Password Checker II is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check each condition for a strong password one by one. This approach is straightforward but inefficient, as it may require multiple passes through the password string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Strong Password Checker II?

Strong Password Checker II uses the following concepts: String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Strong Password Checker II?

Clearly explain your thought process as you write the code. Consider edge cases, such as the minimum password length. Use meaningful variable names to improve code readability.

What are common mistakes when solving Strong Password Checker II?

Forgetting to check for adjacent duplicates while iterating. Not handling the special character check properly.

#2299

Strong Password Checker II

Easy

StringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

We can achieve better performance by checking all conditions in a single pass through the password string. This way, we only need to iterate through the string once, making it more efficient.

⚙️

Algorithm

4 steps

1Step 1: Check if the password length is at least 8 characters.
2Step 2: Initialize flags for lowercase, uppercase, digit, special character, and a variable to track the last character.
3Step 3: Iterate through each character in the password, updating the flags based on the character type and checking for adjacent duplicates.
4Step 4: After the loop, verify if all flags are true.

solution.py19 lines

1def strongPasswordCheckerII(password):
2    if len(password) < 8:
3        return False
4    has_lower = has_upper = has_digit = has_special = False
5    special_characters = "!@#$%^&*()-+"
6    last_char = ''
7    for char in password:
8        if char.islower():
9            has_lower = True
10        elif char.isupper():
11            has_upper = True
12        elif char.isdigit():
13            has_digit = True
14        elif char in special_characters:
15            has_special = True
16        if char == last_char:
17            return False
18        last_char = char
19    return has_lower and has_upper and has_digit and has_special

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the password string. The space complexity is O(1) since we only use a fixed number of variables.

1Checking all conditions in a single pass is more efficient than multiple passes.
2Using flags for character types allows for quick validation of password strength.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.