LC355 - Design Twitter

Problem

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

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.

Solution

Time complexity limited by the number of people some user follows, and it would be $O(n)$ where $n$ is the number of people that person follows.

def __init__(self):
	self.num = 0
	# Default dict doesn't have to be intialized when new ones are made
	self.tweetMap = defaultdict(list)
	self.followMap = defaultdict(set)

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

def getNewsFeed(self, userId: int) -> List[int]:
	feed = []
	minHeap = []
	# user counts as their own follower
	self.followMap[userId].add(userId)
	for followeeId in self.followMap[userId]:
		if followeeId in self.tweetMap:
			index = len(self.tweetMap[followeeId]) - 1
			num, tweetId = self.tweetMap[followeeId][index]
			minHeap.append([num, tweetId, followeeId, index - 1])
	heapq.heapify(minHeap)
	while minHeap and len(feed) < 10:
		num, tweetId, followeeId, index = heapq.heappop(minHeap)
		feed.append(tweetId)
		# if there still are tweets
		if index >= 0:
			num, tweetId = self.tweetMap[followeeId][index]
			heapq.heappush(minHeap, [num, tweetId, followeeId, index-1])

	return feed

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

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