How do you solve Strong Password Checker on LeetCode?

Strong Password Checker (LeetCode #420) 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: Understanding the conditions for a strong password is crucial for efficiently checking and modifying it.

What is the brute force solution for Strong Password Checker?

The brute force approach for Strong Password Checker is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks all conditions for a strong password by iterating through the string and counting the necessary changes. It is straightforward but inefficient, as it may require multiple passes through the string. The optimal Optimal Solution approach improves this to O(n).

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

Strong Password Checker uses the following concepts: String, Greedy, Heap (Priority Queue). The recommended patterns to study are: Greedy algorithms for making optimal replacements., Character counting for checking types and repeats..

What are the interview tips for Strong Password Checker?

Clearly explain your thought process and the conditions for a strong password as you work through the problem. Consider edge cases and how they affect your solution, especially with varying password lengths. Practice breaking down the problem into smaller parts to simplify your approach.

What are common mistakes when solving Strong Password Checker?

Failing to account for all character types when checking password strength. Not considering the impact of excess characters when the password exceeds the maximum length.

#420

Strong Password Checker

Hard

StringGreedyHeap (Priority Queue)Greedy algorithms for making optimal replacements.Character counting for checking types and repeats.

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)

The optimal approach efficiently checks the password's length and character types while managing replacements for repeating characters. It minimizes operations by addressing excess characters directly and adjusting counts based on the password's length.

⚙️

Algorithm

3 steps

1Step 1: Count the number of lowercase letters, uppercase letters, digits, and the number of repeating character sequences.
2Step 2: Determine how many characters need to be added or removed based on the length of the password.
3Step 3: Calculate the total number of replacements needed to meet all conditions and return the total adjustments.

solution.py23 lines

1def strongPasswordChecker(password):
2    n = len(password)
3    has_lower = any(c.islower() for c in password)
4    has_upper = any(c.isupper() for c in password)
5    has_digit = any(c.isdigit() for c in password)
6    missing_types = 3 - (has_lower + has_upper + has_digit)
7    repeat_count = 0
8    i = 2
9    while i < n:
10        if password[i] == password[i - 1] == password[i - 2]:
11            repeat_count += 1
12            i += 2
13        else:
14            i += 1
15    if n < 6:
16        return max(6 - n, missing_types)
17    elif n <= 20:
18        return max(repeat_count, missing_types)
19    else:
20        excess = n - 20
21        replacements = excess
22        repeat_count -= min(repeat_count, excess)
23        return replacements + max(repeat_count, missing_types)

ℹ

Complexity note: The optimal solution runs in O(n) time because it only requires a single pass through the password to check conditions, making it efficient for larger inputs.

1Understanding the conditions for a strong password is crucial for efficiently checking and modifying it.
2Handling edge cases, such as passwords that are too short or too long, is essential for a comprehensive solution.

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