What is the time complexity of Design Twitter?

The optimal solution has a time complexity of O(f*log(u)).

What is the space complexity of Design Twitter?

The space complexity is O(u+t).

What data structures are used to solve Design Twitter?

This problem uses Hash Table, Linked List, Design, Heap (Priority Queue). Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Design Twitter?

Insight 1: We need to maintain the order of tweets to display the most recent ones first. Insight 2: Using a min-heap (priority queue) to merge the sorted lists of tweets from followed users is efficient for getting the top 10 most recent tweets.

Which companies ask LeetCode 355. Design Twitter in interviews?

This problem is frequently asked at Coupang, PayPal, X.

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Design Twitter - LeetCode 355 Solution

Design Twitter - Complete Solution Guide

Design Twitter is LeetCode problem 355, 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

Design a simplified version of Twitter where users can post tweets, follow/unfollow another user, and is able to see the 10 most recent tweets in the user's news feed. Implement the Twitter class: Twitter() Initializes your twitter object. void postTweet(int userId, int tweetId) Composes a new tweet with ID tweetId by the user userId . Each call to this function will be made with a unique tweetId . List<Integer> getNewsFeed(int userId) Retrieves the 10 most recent tweet IDs in the user's news fe

Detailed Explanation

The problem asks us to design a simplified Twitter class that allows users to post tweets, follow/unfollow other users, and retrieve the 10 most recent tweets in a user's news feed. The news feed should include tweets from the user themselves and the users they follow, ordered from most recent to least recent. Each tweet has a unique ID. We need to implement `postTweet`, `getNewsFeed`, `follow`, and `unfollow` methods.

Solution Approach

The solution uses a combination of hash tables (maps) and a min-heap to implement the Twitter functionality. Each user has a list of tweets stored with timestamps. The `getNewsFeed` method retrieves tweets from the user and their followees, merging them using a min-heap to get the 10 most recent tweets. The min-heap stores (timestamp, tweetId, userId, nextTweetIndex) tuples to efficiently track the next most recent tweet for each user. Timestamps are negative to make the min-heap behave like a max-heap for recent tweets.

Step-by-Step Algorithm

Step 1: Initialize the Twitter class with a timestamp counter, a map to store user tweets (userId -> list of (timestamp, tweetId)), and a map to store followees (followerId -> set of followeeIds).
Step 2: `postTweet(userId, tweetId)`: Decrement the timestamp counter (making it negative to represent recency), and append the (timestamp, tweetId) tuple to the user's tweet list.
Step 3: `getNewsFeed(userId)`: Create a set of relevant user IDs (the user themselves and their followees).
Step 4: Initialize a min-heap (priority queue) to store the most recent tweet from each relevant user. The heap stores tuples of (timestamp, tweetId, userId, tweetIndex).
Step 5: Iterate through the relevant users. For each user, if they have tweets, add their most recent tweet to the min-heap.
Step 6: Repeatedly extract the smallest (most recent) tuple from the heap (up to 10 tweets). Add the tweetId to the news feed list.
Step 7: If the user who posted the extracted tweet has more tweets, add their next most recent tweet to the heap.
Step 8: Return the news feed list.
Step 9: `follow(followerId, followeeId)`: Add the followeeId to the followerId's followee set, after validating that the user is not trying to follow themselves.
Step 10: `unfollow(followerId, followeeId)`: Remove the followeeId from the followerId's followee set.

Key Insights

Insight 1: We need to maintain the order of tweets to display the most recent ones first.
Insight 2: Using a min-heap (priority queue) to merge the sorted lists of tweets from followed users is efficient for getting the top 10 most recent tweets.
Insight 3: We should avoid users following themselves, even though the problem statement constraints guarantee that condition will not occur, it's good practice.

Complexity Analysis

Time Complexity: O(f*log(u))

Space Complexity: O(u+t)

Topics

This problem involves: Hash Table, Linked List, Design, Heap (Priority Queue).

Companies

Asked at: Coupang, PayPal, X.

Design Twitter | LeetCode 355

Medium

Hash Table Linked List Design Heap (Priority Queue)

Time: O(f*log(u))Space: O(u+t)

Solution Code

Available in:

import java.util.*;

class Twitter {

    private int time;
    private Map<Integer, List<int[]>> userTweets;
    private Map<Integer, Set<Integer>> following;

    public Twitter() {
        this.time = 0;
        this.userTweets = new HashMap<>();
        this.following = new HashMap<>();
    }
    
    public void postTweet(int userId, int tweetId) {
        this.time -= 1;
        if (!this.userTweets.containsKey(userId)) {
            this.userTweets.put(userId, new ArrayList<>());
        }
        this.userTweets.get(userId).add(new int[]{this.time, tweetId});
    }
    
    public List<Integer> getNewsFeed(int userId) {
        List<Integer> newsFeed = new ArrayList<>();
        PriorityQueue<int[]> minHeap = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
        
        // Create a set of users to pull tweets from: the user themselves and their followees.
        Set<Integer> relevantUserIds = new HashSet<>();
        if (this.following.containsKey(userId)) {
            relevantUserIds.addAll(this.following.get(userId));
        }
        relevantUserIds.add(userId);
        
        // Initialize the heap with the most recent tweet from each relevant user.
        for (int uId : relevantUserIds) {
            if (this.userTweets.containsKey(uId)) {
                List<int[]> tweets = this.userTweets.get(uId);
                int lastTweetIndex = tweets.size() - 1;
                int time = tweets.get(lastTweetIndex)[0];
                int tId = tweets.get(lastTweetIndex)[1];
                // Heap stores (time, tweet_id, user_id, index_of_next_tweet_for_user)
                minHeap.offer(new int[]{time, tId, uId, lastTweetIndex - 1});
            }
        }
        
        // Repeatedly extract the most recent tweet from the heap until we have 10,
        // or the heap is empty.
        while (!minHeap.isEmpty() && newsFeed.size() < 10) {
            int[] top = minHeap.poll();
            int time = top[0];
            int tId = top[1];
            int uId = top[2];
            int nextTweetIndex = top[3];
            newsFeed.add(tId);
            
            // If the user who posted the last tweet has more tweets, add their
            // next most recent one to the heap.
            if (nextTweetIndex >= 0) {
                int nextTime = this.userTweets.get(uId).get(nextTweetIndex)[0];
                int nextTId = this.userTweets.get(uId).get(nextTweetIndex)[1];
                minHeap.offer(new int[]{nextTime, nextTId, uId, nextTweetIndex - 1});
            }
        }
        
        return newsFeed;
    }
    
    public void follow(int followerId, int followeeId) {
        if (followerId != followeeId) {
            if (!this.following.containsKey(followerId)) {
                this.following.put(followerId, new HashSet<>());
            }
            this.following.get(followerId).add(followeeId);
        }
    }
    
    public void unfollow(int followerId, int followeeId) {
        if (this.following.containsKey(followerId)) {
            this.following.get(followerId).remove(followeeId);
        }
    }
}

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Problem Info

Description

Problem 355: Design Twitter - Detailed explanation and solution approaches.

Topics

Hash Table Linked List Design Heap (Priority Queue)

Companies

CoupangPayPalX

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:355

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

10 steps

1
Step 1: Initialize the Twitter class with a timestamp counter, a map to store user tweets (userId -> list of (timestamp, tweetId)), and a map to store followees (followerId -> set of followeeIds).
2
Step 2: `postTweet(userId, tweetId)`: Decrement the timestamp counter (making it negative to represent recency), and append the (timestamp, tweetId) tuple to the user's tweet list.
3
Step 3: `getNewsFeed(userId)`: Create a set of relevant user IDs (the user themselves and their followees).
4
Step 4: Initialize a min-heap (priority queue) to store the most recent tweet from each relevant user. The heap stores tuples of (timestamp, tweetId, userId, tweetIndex).
5
Step 5: Iterate through the relevant users. For each user, if they have tweets, add their most recent tweet to the min-heap.
6
Step 6: Repeatedly extract the smallest (most recent) tuple from the heap (up to 10 tweets). Add the tweetId to the news feed list.
7
Step 7: If the user who posted the extracted tweet has more tweets, add their next most recent tweet to the heap.
8
Step 8: Return the news feed list.
9
Step 9: `follow(followerId, followeeId)`: Add the followeeId to the followerId's followee set, after validating that the user is not trying to follow themselves.
10
Step 10: `unfollow(followerId, followeeId)`: Remove the followeeId from the followerId's followee set.

💡

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(f*log(u))

💾

Space Complexity

O(u+t)

📝

Why?

The time complexity of `getNewsFeed` is dominated by the heap operations. The heap size is bounded by the number of users the given user follows, and each heap operation takes logarithmic time. The space complexity is due to storing tweets and followee relationships, which grows linearly with the number of users and tweets.

📖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

Merge k Sorted Lists

Treat the tweets of each followed user as a sorted list and merge them using a merge k sorted lists algorithm. This is conceptually similar to the heap approach.

Time: O(N log k), where N is the total number of tweets across all lists and k is the number of lists being merged (number of followees). In this case, could be O(t * log(f)) where t is the total number of tweets to get and f is the number of followers.Space: O(k), where k is the number of lists being merged. In the specific case, this would be O(f) where f is the number of followers.

Trade-offs: The heap approach is generally preferred because it's easier to implement, and in practice, the difference in performance is likely to be negligible. Merge k sorted lists can be more efficient in certain cases when the heap operations are a bottleneck, especially with very large number of followees, but is harder to implement.

✨

Key Insights

1Insight 1: We need to maintain the order of tweets to display the most recent ones first.
2Insight 2: Using a min-heap (priority queue) to merge the sorted lists of tweets from followed users is efficient for getting the top 10 most recent tweets.
3Insight 3: We should avoid users following themselves, even though the problem statement constraints guarantee that condition will not occur, it's good practice.

📚

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: User has no tweets. The code correctly handles this by checking `if uId in self.user_tweets` and continuing to the next user.
Edge case 2: User follows no one. The news feed will only contain the user's own tweets.
Edge case 3: User has fewer than 10 tweets in their news feed. The `while` loop in `getNewsFeed` terminates when the heap is empty or 10 tweets have been collected, ensuring that fewer than 10 tweets are returned when necessary.
Edge case 4: FollowerId and followeeId are the same in follow/unfollow. The code includes `if followerId != followeeId` to prevent self-following which can lead to errors or infinite loops.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where a user has no tweets or no followers. This can lead to errors when trying to access non-existent data.

✗Common mistake 2: Not properly ordering the tweets in the news feed. The most recent tweets must come first.

✗Common mistake 3: Forgetting to consider the user's own tweets when constructing the news feed. The user's tweets should always be included.

✗Common mistake 4: Not handling duplicate followers which would slow down the performance if adding the same tweets twice from the same user

✅

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: Social media feeds (Twitter, Facebook, Instagram) use similar algorithms to display relevant content to users.

🏦Real-world scenario 2: News aggregators and content recommendation systems leverage these techniques to prioritize and display the most timely and relevant information.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 23, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

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

@InterviewProJun 29

❓

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 Twitter"?

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?