How do you solve Design Authentication Manager on LeetCode?

Design Authentication Manager (LeetCode #1797) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a HashMap allows for efficient token management and quick access to token expiry times.

What is the brute force solution for Design Authentication Manager?

The brute force approach for Design Authentication Manager is Brute Force, with O(n²) time complexity and O(n) space complexity. In the brute force approach, we can use a simple list to store tokens and their expiry times. For each operation, we will iterate through the list to check for expired tokens and perform the required actions. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Design Authentication Manager?

Design Authentication Manager uses the following concepts: Hash Table, Linked List, Design, Doubly-Linked List. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Design Authentication Manager?

Always clarify the requirements and edge cases before diving into coding. Think about the data structures that can optimize your solution. Explain your thought process clearly while coding to demonstrate your understanding.

What are common mistakes when solving Design Authentication Manager?

Failing to check if a token exists before trying to renew it. Not correctly updating the expiry time when renewing a token.

#1797

Design Authentication Manager

Medium

Hash TableLinked ListDesignDoubly-Linked ListHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a HashMap to store tokens along with their expiry times, allowing for O(1) access to check and update tokens. This significantly reduces the time complexity for renewals and counts.

⚙️

Algorithm

4 steps

1Step 1: Use a HashMap to store tokenId as the key and its expiry time as the value.
2Step 2: For 'generate', simply add the tokenId and its expiry time to the HashMap.
3Step 3: For 'renew', check if the tokenId exists and if it is unexpired; if so, update its expiry time.
4Step 4: For 'countUnexpiredTokens', iterate through the HashMap and count tokens that are still valid.

solution.py14 lines

1class AuthenticationManager:
2    def __init__(self, timeToLive: int):
3        self.timeToLive = timeToLive
4        self.tokens = {}
5
6    def generate(self, tokenId: str, currentTime: int) -> None:
7        self.tokens[tokenId] = currentTime + self.timeToLive
8
9    def renew(self, tokenId: str, currentTime: int) -> None:
10        if tokenId in self.tokens and self.tokens[tokenId] > currentTime:
11            self.tokens[tokenId] = currentTime + self.timeToLive
12
13    def countUnexpiredTokens(self, currentTime: int) -> int:
14        return sum(1 for expiry in self.tokens.values() if expiry > currentTime)

ℹ

Complexity note: The time complexity is O(n) for counting unexpired tokens because we may need to check each token once. The space complexity is O(n) due to storing tokens in a HashMap.

1Using a HashMap allows for efficient token management and quick access to token expiry times.
2Understanding the implications of token expiration timing is crucial for implementing renewals correctly.

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