LeetCode #355 — MEDIUM

Design Twitter

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

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 feed. Each item in the news feed must be posted by users who the user followed or by the user themself. Tweets must be ordered from most recent to least recent.
void follow(int followerId, int followeeId) The user with ID followerId started following the user with ID followeeId.
void unfollow(int followerId, int followeeId) The user with ID followerId started unfollowing the user with ID followeeId.

Example 1:

Input
["Twitter", "postTweet", "getNewsFeed", "follow", "postTweet", "getNewsFeed", "unfollow", "getNewsFeed"]
[[], [1, 5], [1], [1, 2], [2, 6], [1], [1, 2], [1]]
Output
[null, null, [5], null, null, [6, 5], null, [5]]

Explanation
Twitter twitter = new Twitter();
twitter.postTweet(1, 5); // User 1 posts a new tweet (id = 5).
twitter.getNewsFeed(1);  // User 1's news feed should return a list with 1 tweet id -> [5]. return [5]
twitter.follow(1, 2);    // User 1 follows user 2.
twitter.postTweet(2, 6); // User 2 posts a new tweet (id = 6).
twitter.getNewsFeed(1);  // User 1's news feed should return a list with 2 tweet ids -> [6, 5]. Tweet id 6 should precede tweet id 5 because it is posted after tweet id 5.
twitter.unfollow(1, 2);  // User 1 unfollows user 2.
twitter.getNewsFeed(1);  // User 1's news feed should return a list with 1 tweet id -> [5], since user 1 is no longer following user 2.

Constraints:

1 <= userId, followerId, followeeId <= 500
0 <= tweetId <= 10⁴
All the tweets have unique IDs.
At most 3 * 10⁴ calls will be made to postTweet, getNewsFeed, follow, and unfollow.
A user cannot follow himself.

Step 01

Brute Force Baseline

Problem summary: 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 feed. Each item in the news feed must be posted by users who the user followed or by the user themself. Tweets must be ordered from most recent to least recent. void follow(int followerId, int followeeId) The user with ID followerId started following the user with ID followeeId. void unfollow(int followerId, int followeeId) The user with ID followerId started unfollowing the user with ID followeeId.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Linked List · Design

Example 1

["Twitter","postTweet","getNewsFeed","follow","postTweet","getNewsFeed","unfollow","getNewsFeed"]
[[],[1,5],[1],[1,2],[2,6],[1],[1,2],[1]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #355: Design Twitter
class Twitter {
    private Map<Integer, List<Integer>> userTweets;
    private Map<Integer, Set<Integer>> userFollowing;
    private Map<Integer, Integer> tweets;
    private int time;

    /** Initialize your data structure here. */
    public Twitter() {
        userTweets = new HashMap<>();
        userFollowing = new HashMap<>();
        tweets = new HashMap<>();
        time = 0;
    }

    /** Compose a new tweet. */
    public void postTweet(int userId, int tweetId) {
        userTweets.computeIfAbsent(userId, k -> new ArrayList<>()).add(tweetId);
        tweets.put(tweetId, ++time);
    }

    /**
     * Retrieve the 10 most recent tweet ids in the user's news feed. Each item in the news feed
     * must be posted by users who the user followed or by the user herself. Tweets must be ordered
     * from most recent to least recent.
     */
    public List<Integer> getNewsFeed(int userId) {
        Set<Integer> following = userFollowing.getOrDefault(userId, new HashSet<>());
        Set<Integer> users = new HashSet<>(following);
        users.add(userId);
        PriorityQueue<Integer> pq
            = new PriorityQueue<>(10, (a, b) -> (tweets.get(b) - tweets.get(a)));
        for (Integer u : users) {
            List<Integer> userTweet = userTweets.get(u);
            if (userTweet != null && !userTweet.isEmpty()) {
                for (int i = userTweet.size() - 1, k = 10; i >= 0 && k > 0; --i, --k) {
                    pq.offer(userTweet.get(i));
                }
            }
        }
        List<Integer> res = new ArrayList<>();
        while (!pq.isEmpty() && res.size() < 10) {
            res.add(pq.poll());
        }
        return res;
    }

    /** Follower follows a followee. If the operation is invalid, it should be a no-op. */
    public void follow(int followerId, int followeeId) {
        userFollowing.computeIfAbsent(followerId, k -> new HashSet<>()).add(followeeId);
    }

    /** Follower unfollows a followee. If the operation is invalid, it should be a no-op. */
    public void unfollow(int followerId, int followeeId) {
        userFollowing.computeIfAbsent(followerId, k -> new HashSet<>()).remove(followeeId);
    }
}

/**
 * Your Twitter object will be instantiated and called as such:
 * Twitter obj = new Twitter();
 * obj.postTweet(userId,tweetId);
 * List<Integer> param_2 = obj.getNewsFeed(userId);
 * obj.follow(followerId,followeeId);
 * obj.unfollow(followerId,followeeId);
 */

// Accepted solution for LeetCode #355: Design Twitter
type Tweet struct {
    count int
    tweetId int
    followeeId int
    index int
}

type TweetHeap []*Tweet
func (h TweetHeap) Len() int           { return len(h) }
func (h TweetHeap) Less(i, j int) bool { return h[i].count < h[j].count }
func (h TweetHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *TweetHeap) Push(x interface{}) {*h = append(*h, x.(*Tweet))}
func (h *TweetHeap) Pop() interface{} {
	old := *h
	n := len(old)
	x := old[n-1]
	*h = old[0 : n-1]
	return x
}

type Twitter struct {
    count int
    tweetMap map[int][]*Tweet
    followMap map[int]map[int]bool
}

func Constructor() Twitter {
    return Twitter{tweetMap: make(map[int][]*Tweet), followMap: make(map[int]map[int]bool)}
}

func (this *Twitter) PostTweet(userId int, tweetId int)  {
    if _, ok := this.tweetMap[userId]; !ok {
        this.tweetMap[userId] = make([]*Tweet, 0)
    }
    this.tweetMap[userId] = append(this.tweetMap[userId], &Tweet{count: this.count, tweetId: tweetId})
    this.count -= 1
}


func (this *Twitter) GetNewsFeed(userId int) []int {
    res := make([]int, 0)
    minHeap := TweetHeap{}
    
    if _, ok := this.followMap[userId]; !ok {
        this.followMap[userId] = make(map[int]bool)
    }
    this.followMap[userId][userId] = true;
    for followeeId, _ := range this.followMap[userId] {
        if _, ok := this.tweetMap[followeeId]; ok {
            index := len(this.tweetMap[followeeId]) - 1
            tweet := this.tweetMap[followeeId][index]
            heap.Push(&minHeap, &Tweet{
                count: tweet.count, 
                tweetId: tweet.tweetId, 
                followeeId: followeeId, 
                index: index - 1})
        }
    }
    
    for len(minHeap) > 0 && len(res) < 10 {
        tweet := heap.Pop(&minHeap).(*Tweet)
        res = append(res, tweet.tweetId)
        if tweet.index >= 0 {
            nextTweet := this.tweetMap[tweet.followeeId][tweet.index]
            heap.Push(&minHeap, &Tweet{count: nextTweet.count, 
                                     tweetId: nextTweet.tweetId,
                                     followeeId: tweet.followeeId,
                                     index: tweet.index - 1})
        }
    }
    return res
}


func (this *Twitter) Follow(followerId int, followeeId int)  {
    if _, ok := this.followMap[followerId]; !ok {
        this.followMap[followerId] = make(map[int]bool)
    }
    this.followMap[followerId][followeeId] = true
}


func (this *Twitter) Unfollow(followerId int, followeeId int)  {
    if _, ok := this.followMap[followerId]; ok {
        delete(this.followMap[followerId], followeeId)
    }
}

# Accepted solution for LeetCode #355: Design Twitter
class Twitter:
    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.user_tweets = defaultdict(list)
        self.user_following = defaultdict(set)
        self.tweets = defaultdict()
        self.time = 0

    def postTweet(self, userId: int, tweetId: int) -> None:
        """
        Compose a new tweet.
        """
        self.time += 1
        self.user_tweets[userId].append(tweetId)
        self.tweets[tweetId] = self.time

    def getNewsFeed(self, userId: int) -> List[int]:
        """
        Retrieve the 10 most recent tweet ids in the user's news feed. Each item in the news feed must be posted by users who the user followed or by the user herself. Tweets must be ordered from most recent to least recent.
        """
        following = self.user_following[userId]
        users = set(following)
        users.add(userId)
        tweets = [self.user_tweets[u][::-1][:10] for u in users]
        tweets = sum(tweets, [])
        return nlargest(10, tweets, key=lambda tweet: self.tweets[tweet])

    def follow(self, followerId: int, followeeId: int) -> None:
        """
        Follower follows a followee. If the operation is invalid, it should be a no-op.
        """
        self.user_following[followerId].add(followeeId)

    def unfollow(self, followerId: int, followeeId: int) -> None:
        """
        Follower unfollows a followee. If the operation is invalid, it should be a no-op.
        """
        following = self.user_following[followerId]
        if followeeId in following:
            following.remove(followeeId)


# Your Twitter object will be instantiated and called as such:
# obj = Twitter()
# obj.postTweet(userId,tweetId)
# param_2 = obj.getNewsFeed(userId)
# obj.follow(followerId,followeeId)
# obj.unfollow(followerId,followeeId)

// Accepted solution for LeetCode #355: Design Twitter
use std::cmp::Reverse;
use std::collections::BinaryHeap;
use std::collections::HashMap;
use std::collections::HashSet;

type Tweet = (Reverse<usize>, i32);

#[derive(Default)]
struct Twitter {
    time: usize,
    users: HashMap<i32, HashSet<i32>>,
    tweets: HashMap<i32, Vec<Tweet>>,
    limit: usize,
}

impl Twitter {
    fn new() -> Self {
        let time = 0;
        let users = HashMap::new();
        let tweets = HashMap::new();
        let limit = 10;
        Twitter {
            time,
            users,
            tweets,
            limit,
        }
    }

    fn post_tweet(&mut self, user_id: i32, tweet_id: i32) {
        self.time += 1;
        let tweet: Tweet = (Reverse(self.time), tweet_id);
        self.tweets.entry(user_id).or_default().push(tweet);
    }

    fn get_news_feed(&mut self, user_id: i32) -> Vec<i32> {
        let mut pq: BinaryHeap<Tweet> = BinaryHeap::with_capacity(self.limit + 1);
        let mut res: Vec<i32> = vec![];
        let followers = self.users.entry(user_id).or_default();
        followers.insert(user_id);
        for &user in followers.iter() {
            let tweets = self.tweets.entry(user).or_default();
            for tweet in tweets {
                pq.push(*tweet);
                if pq.len() > self.limit {
                    pq.pop();
                }
            }
        }
        while !pq.is_empty() {
            let earliest = pq.pop().unwrap();
            res.push(earliest.1);
        }
        res.reverse();
        res
    }

    fn follow(&mut self, follower_id: i32, followee_id: i32) {
        self.users
            .entry(follower_id)
            .or_default()
            .insert(followee_id);
    }

    fn unfollow(&mut self, follower_id: i32, followee_id: i32) {
        self.users
            .entry(follower_id)
            .or_default()
            .remove(&followee_id);
    }
}

#[test]
fn test() {
    let mut twitter = Twitter::new();
    twitter.post_tweet(1, 5);
    assert_eq!(twitter.get_news_feed(1), vec![5]);
    twitter.follow(1, 2);
    twitter.post_tweet(2, 6);
    assert_eq!(twitter.get_news_feed(1), vec![6, 5]);
    twitter.unfollow(1, 2);
    assert_eq!(twitter.get_news_feed(1), vec![5]);
}

// Accepted solution for LeetCode #355: Design Twitter
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #355: Design Twitter
// class Twitter {
//     private Map<Integer, List<Integer>> userTweets;
//     private Map<Integer, Set<Integer>> userFollowing;
//     private Map<Integer, Integer> tweets;
//     private int time;
// 
//     /** Initialize your data structure here. */
//     public Twitter() {
//         userTweets = new HashMap<>();
//         userFollowing = new HashMap<>();
//         tweets = new HashMap<>();
//         time = 0;
//     }
// 
//     /** Compose a new tweet. */
//     public void postTweet(int userId, int tweetId) {
//         userTweets.computeIfAbsent(userId, k -> new ArrayList<>()).add(tweetId);
//         tweets.put(tweetId, ++time);
//     }
// 
//     /**
//      * Retrieve the 10 most recent tweet ids in the user's news feed. Each item in the news feed
//      * must be posted by users who the user followed or by the user herself. Tweets must be ordered
//      * from most recent to least recent.
//      */
//     public List<Integer> getNewsFeed(int userId) {
//         Set<Integer> following = userFollowing.getOrDefault(userId, new HashSet<>());
//         Set<Integer> users = new HashSet<>(following);
//         users.add(userId);
//         PriorityQueue<Integer> pq
//             = new PriorityQueue<>(10, (a, b) -> (tweets.get(b) - tweets.get(a)));
//         for (Integer u : users) {
//             List<Integer> userTweet = userTweets.get(u);
//             if (userTweet != null && !userTweet.isEmpty()) {
//                 for (int i = userTweet.size() - 1, k = 10; i >= 0 && k > 0; --i, --k) {
//                     pq.offer(userTweet.get(i));
//                 }
//             }
//         }
//         List<Integer> res = new ArrayList<>();
//         while (!pq.isEmpty() && res.size() < 10) {
//             res.add(pq.poll());
//         }
//         return res;
//     }
// 
//     /** Follower follows a followee. If the operation is invalid, it should be a no-op. */
//     public void follow(int followerId, int followeeId) {
//         userFollowing.computeIfAbsent(followerId, k -> new HashSet<>()).add(followeeId);
//     }
// 
//     /** Follower unfollows a followee. If the operation is invalid, it should be a no-op. */
//     public void unfollow(int followerId, int followeeId) {
//         userFollowing.computeIfAbsent(followerId, k -> new HashSet<>()).remove(followeeId);
//     }
// }
// 
// /**
//  * Your Twitter object will be instantiated and called as such:
//  * Twitter obj = new Twitter();
//  * obj.postTweet(userId,tweetId);
//  * List<Integer> param_2 = obj.getNewsFeed(userId);
//  * obj.follow(followerId,followeeId);
//  * obj.unfollow(followerId,followeeId);
//  */

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

COPY TO ARRAY

O(n) time

O(n) space

Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.

IN-PLACE POINTERS

O(n) time

O(1) space

Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.

Shortcut: Traverse once + re-wire pointers → O(n) time, O(1) space. Dummy head nodes simplify edge cases.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Losing head/tail while rewiring

Wrong move: Pointer updates overwrite references before they are saved.

Usually fails on: List becomes disconnected mid-operation.

Fix: Store next pointers first and use a dummy head for safer joins.