LC 355 - Design Twitter

Question

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.

Links

Question here and solution here

Solution

concept

The main thing here is to define the proper data structure for each function properly. And we need to define a time variable to keep track the timestamp. We will use a heap to pop the most k recent tweets

code

class Twitter:

    def __init__(self):
        self.time = 0
        self.follow_map = defaultdict(set) # followerID: (foloweeIDs)
        self.tweet_map = defaultdict(list) # userID: [(time,tweetID)]

    def postTweet(self, userId: int, tweetId: int) -> None:
        self.tweet_map[userId].append((self.time, tweetId))
        self.time -= 1

    def getNewsFeed(self, userId: int) -> List[int]:
        recent = []
        min_heap = []
        
        self.follow_map[userId].add(userId)
        for followeeId in self.follow_map[userId]:
            min_heap.extend(self.tweet_map[followeeId])

        heapq.heapify(min_heap)
        print(min_heap)
        while min_heap and len(recent) < 10:
            _, tweetId = heapq.heappop(min_heap)
            recent.append(tweetId)
        return recent

    def follow(self, followerId: int, followeeId: int) -> None:
        self.follow_map[followerId].add(followeeId)

    def unfollow(self, followerId: int, followeeId: int) -> None:
        if followeeId in self.follow_map[followerId]:
            self.follow_map[followerId].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)

class Twitter:
    """
    NeetSolution
    Abit more optimised for the heap
    time complexity: O(logn) for heap operations
    overall time complexity: O(k)
    """

    def __init__(self):
        self.count = 0
        # userId -> list of [count, tweetId]
        self.tweet_map = defaultdict(list)
        # userId -> set of followerId
        self.follow_map = defaultdict(set)
    
    def postTweet(self, userId: int, tweetId: int) -> None:
        self.tweet_map[userId].append([self.count, tweetId])
        # since we need minHeap
        self.count -= 1
    
    def getNewsFeed(self, userId: int) -> List[int]:
        result = []
        minHeap = []
        # user follow themselves too
        self.follow_map[userId].add(userId)

        for followeeId in self.follow_map[userId]:
            if followeeId in self.tweet_map:
                index = len(self.tweet_map[followeeId]) - 1
                count, tweetId = self.tweet_map[followeeId][index]
                heapq.heappush(minHeap, [count, tweetId, followeeId, index - 1])
        
        while minHeap and len(result) < 10:
            count, tweetId, followeeId, index = heapq.heappop(minHeap)
            result.append(tweetId)
            if index >= 0:
                count, tweetId = self.tweet_map[followeeId][index]
                heapq.heappush(minHeap, [count, tweetId, followeeId, index - 1])

        return result
        
    def follow(self, followerId: int, followeeId: int) -> None:
        self.follow_map[followerId].add(followeeId)

    def unfollow(self, followerId: int, followeeId: int) -> None:
        if followeeId in self.follow_map[followerId]:
            self.follow_map[followerId].remove(followeeId)

Complexity

time: $O(n \log n)$
space: $O(n)$