from heapq import nlargest
class Twitter:
    '''
    Time Complexity: O(n)
    Since, O(n log 10) will reduce to O(n)
    Space Complexity: O(n)
    where, n is total number of posts
    Solved in 37 mins, all by yourself! Good job!
    NOTE: You missed the point that for getNewsFeed you have to also get the posts created by user, 
    not just the posts from users they follow! This wasted a lot of time!
    '''
    def __init__(self):
        self.following = dict()
        self.posts = dict()
        self.time = 0

    def postTweet(self, userId: int, tweetId: int) -> None:
        self.time += 1
        self.posts.setdefault(userId, list()).append((self.time, tweetId))

    def getNewsFeed(self, userId: int) -> List[int]:
        feed = []
        for user in self.following.get(userId, [userId]):
            feed += self.posts.get(user, [])
        return [tweet for _, tweet in nlargest(10, feed)]

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

    def unfollow(self, followerId: int, followeeId: int) -> None:
        self.following.get(followerId, set()).discard(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)