What is the time complexity of Design Authentication Manager?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Design Authentication Manager?

This problem uses Hash Table, Linked List, Design, Doubly-Linked List. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Design Authentication Manager?

Key Insight 1: Using a hash table (dictionary/unordered_map) to store the tokens allows efficient lookups, insertions, and deletions based on the `tokenId`. Key Insight 2: The 'currentTime' is strictly increasing, simplifying the problem since we don't need to sort or handle out-of-order events. We only need to check if a token's expiry time is greater than the current time.

Which companies ask LeetCode 1797. Design Authentication Manager in interviews?

This problem is frequently asked at Confluent, Docusign, Genpact, X.

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Design Authentication Manager - LeetCode 1797 Solution

Design Authentication Manager - Complete Solution Guide

Design Authentication Manager is LeetCode problem 1797, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There is an authentication system that works with authentication tokens. For each session, the user will receive a new authentication token that will expire timeToLive seconds after the currentTime . If the token is renewed, the expiry time will be extended to expire timeToLive seconds after the (potentially different) currentTime . Implement the AuthenticationManager class: AuthenticationManager(int timeToLive) constructs the AuthenticationManager and sets the timeToLive . generate(string token

Detailed Explanation

The problem asks us to design an authentication system that manages user authentication tokens. Each token has a `tokenId` and an expiration time, which is determined by the `timeToLive` and the `currentTime` when the token is generated or renewed. The system needs to support generating new tokens, renewing existing unexpired tokens, and counting the number of unexpired tokens at a given time. The `currentTime` is strictly increasing across all operations, and it's important to consider that if a token expires at time t, and another action happens at time t, the expiration takes precedence. This implies checking the validity of a token *before* performing any operation on it.

Solution Approach

The provided solution uses a hash table (`tokens`) to store the `tokenId` and its corresponding expiration time. The `generate` function simply adds a new entry to the hash table. The `renew` function checks if the token exists and if its expiry time is greater than the current time; if so, it updates the expiry time. The `countUnexpiredTokens` function iterates through the hash table, checks for expired tokens, removes them, and then returns the number of remaining tokens.

Step-by-Step Algorithm

Step 1: Initialize the AuthenticationManager with the given `timeToLive` and an empty hash table `tokens` to store token information.
Step 2: Implement the `generate` function: Store a new token in the `tokens` hash table, with the `tokenId` as the key and the `currentTime + timeToLive` as the value (expiration time).
Step 3: Implement the `renew` function: Check if the `tokenId` exists in the `tokens` hash table AND if the token's expiration time is greater than the `currentTime`. If both conditions are true, update the token's expiration time to `currentTime + timeToLive`.
Step 4: Implement the `countUnexpiredTokens` function: Iterate through the `tokens` hash table. For each token, check if its expiration time is less than or equal to the `currentTime`. If it is, add the `tokenId` to a list of tokens to be removed. After iterating, remove all expired tokens from the `tokens` hash table.
Step 5: Finally, return the number of tokens remaining in the `tokens` hash table. This represents the number of unexpired tokens.

Key Insights

Key Insight 1: Using a hash table (dictionary/unordered_map) to store the tokens allows efficient lookups, insertions, and deletions based on the `tokenId`.
Key Insight 2: The 'currentTime' is strictly increasing, simplifying the problem since we don't need to sort or handle out-of-order events. We only need to check if a token's expiry time is greater than the current time.
Key Insight 3: When counting unexpired tokens, it's necessary to remove expired tokens from the data structure to maintain an accurate count and prevent future operations from acting on them. This removal needs to happen *before* returning the count.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, Linked List, Design, Doubly-Linked List.

Companies

Asked at: Confluent, Docusign, Genpact, X.

Design Authentication Manager | LeetCode 1797

Medium

Hash Table Linked List Design Doubly-Linked List

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.HashMap;
import java.util.ArrayList;

class AuthenticationManager {

    private int timeToLive;
    private HashMap<String, Integer> tokens;

    public AuthenticationManager(int timeToLive) {
        this.timeToLive = timeToLive;
        this.tokens = new HashMap<>();
    }
    
    public void generate(String tokenId, int currentTime) {
        this.tokens.put(tokenId, currentTime + this.timeToLive);
    }
    
    public void renew(String tokenId, int currentTime) {
        Integer expirationTime = this.tokens.get(tokenId);
        if (expirationTime != null && expirationTime > currentTime) {
            this.tokens.put(tokenId, currentTime + this.timeToLive);
        }
    }
    
    public int countUnexpiredTokens(int currentTime) {
        ArrayList<String> toRemove = new ArrayList<>();
        for (String tokenId : this.tokens.keySet()) {
            int expiryTime = this.tokens.get(tokenId);
            if (expiryTime <= currentTime) {
                toRemove.add(tokenId);
            }
        }
        
        for (String tokenId : toRemove) {
            this.tokens.remove(tokenId);
        }
        
        return this.tokens.size();
    }
}

/**
 * Your AuthenticationManager object will be instantiated and called as such:
 * AuthenticationManager obj = new AuthenticationManager(timeToLive);
 * obj.generate(tokenId,currentTime);
 * obj.renew(tokenId,currentTime);
 * int param_3 = obj.countUnexpiredTokens(currentTime);
 */

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Problem Info

Description

Problem 1797: Design Authentication Manager - Detailed explanation and solution approaches.

Topics

Hash Table Linked List Design Doubly-Linked List

Companies

ConfluentDocusignGenpactX

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:1797

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Initialize the AuthenticationManager with the given `timeToLive` and an empty hash table `tokens` to store token information.
2
Step 2: Implement the `generate` function: Store a new token in the `tokens` hash table, with the `tokenId` as the key and the `currentTime + timeToLive` as the value (expiration time).
3
Step 3: Implement the `renew` function: Check if the `tokenId` exists in the `tokens` hash table AND if the token's expiration time is greater than the `currentTime`. If both conditions are true, update the token's expiration time to `currentTime + timeToLive`.
4
Step 4: Implement the `countUnexpiredTokens` function: Iterate through the `tokens` hash table. For each token, check if its expiration time is less than or equal to the `currentTime`. If it is, add the `tokenId` to a list of tokens to be removed. After iterating, remove all expired tokens from the `tokens` hash table.
5
Step 5: Finally, return the number of tokens remaining in the `tokens` hash table. This represents the number of unexpired tokens.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The `generate` and `renew` functions use hash table lookups and insertions, which have an average time complexity of O(1). However, in the worst-case scenario (e.g., hash collisions), they can be O(n). The `countUnexpiredTokens` function must iterate through all tokens to remove expired ones, resulting in O(n) time complexity. The space complexity is determined by the maximum number of tokens stored, which can be up to n.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using a Priority Queue/Min-Heap

Store tokens in a priority queue sorted by expiry time. `countUnexpiredTokens` could then poll from the priority queue until the peek element has expired.

Time: O(log n) for generate and renew (heap insertion/update). O(k log n) for countUnexpiredTokens where k is the number of expired tokens removed.Space: O(n)

Trade-offs: This approach could be more efficient if `countUnexpiredTokens` is called frequently and the number of expired tokens is relatively small. However, the overhead of maintaining the priority queue might outweigh the benefits if insertions/renewals are more frequent than counts.

Using a Sorted List

Maintain a sorted list of tokens based on their expiry time. This allows efficient retrieval of unexpired tokens.

Time: O(n log n) initially to create the sorted list. generate and renew would be O(log n) on average for insertion and update (binary search). countUnexpiredTokens could then stop scanning once the expiration time is greater than current time. It could provide better performance than hashmap in some scenarios.Space: O(n)

Trade-offs: This approach could be more efficient if `countUnexpiredTokens` is called frequently and the tokens are already almost sorted. However, the overhead of maintaining the sorted list might outweigh the benefits if insertions/renewals are more frequent than counts and they add random expiry times into the structure.

✨

Key Insights

1Key Insight 1: Using a hash table (dictionary/unordered_map) to store the tokens allows efficient lookups, insertions, and deletions based on the `tokenId`.
2Key Insight 2: The 'currentTime' is strictly increasing, simplifying the problem since we don't need to sort or handle out-of-order events. We only need to check if a token's expiry time is greater than the current time.
3Key Insight 3: When counting unexpired tokens, it's necessary to remove expired tokens from the data structure to maintain an accurate count and prevent future operations from acting on them. This removal needs to happen *before* returning the count.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Linked List

Elements connected via pointers. Like a chain where each link knows the next.

Edge Cases

Edge case 1: Renewing a non-existent token: The `renew` function correctly handles this by first checking if the `tokenId` exists in the hash table. If it doesn't, the function does nothing, as specified in the problem description.
Edge case 2: Renewing an expired token: The `renew` function checks if the token's expiry time is greater than the current time. If the token has already expired (expiry time <= current time), the function does nothing.
Edge case 3: Calling countUnexpiredTokens with an empty token list: The `countUnexpiredTokens` will correctly return 0 since there are no tokens to iterate through.
Edge case 4: TimeToLive = 1 and currentTime are close. This is handled correctly as the code ensures expiry is checked before renewal or counting

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not removing expired tokens during the `countUnexpiredTokens` operation: Failing to remove expired tokens will lead to an inaccurate count and can cause issues if those tokens are later renewed.

✗Common mistake 2: Not checking if a token exists before renewing it: This can lead to errors or unexpected behavior if the code attempts to access a non-existent token.

✗Common mistake 3: Forgetting that a token expires at time 't' before any other action can be performed at time 't'.

✗Common mistake 4: Not handling the case where the current time has already passed the expiry time of the token during a renew operation. Failing to check this condition before renewing would lead to incorrect expiry times.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Session management in web applications: Authentication tokens are used to maintain user sessions. The system needs to efficiently generate, renew, and validate tokens.

🏦Real-world scenario 2: API key management: APIs often use keys with expiration dates to control access. The system needs to track the validity of these keys and allow them to be renewed.

🎮Real-world scenario 3: OAuth token management: OAuth tokens used to authorize access to resources also have expiry times that must be managed.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 25, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How does the fast/slow pointer technique work?

Fast pointer moves 2 steps while slow moves 1 step. When fast reaches the end, slow is at middle. If there's a cycle, they will meet. This technique is used for: finding middle node, detecting cycles, finding cycle start point, and checking palindromes.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Design Authentication Manager"?

For this medium-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether linked list techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?