Java 164ms solution (97%) using head for merging one tweet at a time instead of sorting the entire tweets

  • 3

    Key points:

    1. Maintaining a sorted tweets list of each user with O(1) time by adding to the tail each time.
    2. Customized a comparator class which traverses a list from its tail to its head.
    3. The heap, first, store all the tail of those tweet lists followed by a user.
    4. Poll the list contains the latest tweet and then moves its index to the next tweet. If this list has next tweet, put it back to the heap.
    public class Twitter {
        Map<Integer, Set<Integer>> follows; // store one user's follows
        Map<Integer, List<Integer>> userTweets; // store index of one user's tweets 
        List<Integer> tweets; // store every tweets and use its index as the sequence id or timestamp 
        // the wrapper class of one user's tweets, which is similar to an itertor to read from backward 
        class UserTweetList{
            List<Integer> list;
            int idx;
            UserTweetList(List<Integer> list) {
                this.list = list;
                if (list == null || list.isEmpty() ) idx = -1;
                else idx = list.size()-1; // initial the index from the tail of a list
            // check the timestamp of current index
            Integer peek(){ 
                if (idx < 0) return null;
                return list.get(idx);
            // get the tweet ID from the tweet list and then move the index to the next one
            Integer getVal(){
                if (idx < 0) return null;
                int id = list.get(idx--); 
                return tweets.get(id);
        // Comparator class for max-heap ordered by timestamp
        class UTLComp implements Comparator<UserTweetList>{ 
            public int compare(UserTweetList ut1, UserTweetList ut2) {
                if (ut1.peek()==null && ut2.peek()==null) return 0;
                if (ut1.peek()==null) return 1;
                if (ut2.peek()==null) return -1;
                return ut2.peek()-ut1.peek();
        /** Initialize your data structure here. */
        public Twitter() {
            follows = new HashMap<Integer, Set<Integer>>();
            userTweets = new HashMap<Integer, List<Integer>>();
            tweets = new ArrayList<Integer>();
        /** Compose a new tweet. */
        public void postTweet(int userId, int tweetId) { // O(1)
            if (! follows.containsKey(userId) ) {
                follows.put(userId, new HashSet<Integer>() );
            if (! userTweets.containsKey(userId) ) { 
                userTweets.put(userId, new ArrayList<Integer>());
            // add to the end of user's tweet list so it is also sorted
        /** 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) {
            List<Integer> ret = new ArrayList<Integer>();
            PriorityQueue<UserTweetList> heap = new PriorityQueue<UserTweetList>(1, new UTLComp() );
            if (! follows.containsKey(userId) ) return ret;
            // gather the following users' list. k*O(k), k is the number of follows
            for(int follow: follows.get(userId)) {   
                if (userTweets.containsKey(follow)) {
                    List<Integer> list = userTweets.get(follow);
                    if (! list.isEmpty() )
                        heap.offer( new UserTweetList(list));
            // similar to merge sort, 
            // take the largest one from a list and then poll the next one in this list into the queue
            while ( (ret.size() < 10) && ! heap.isEmpty() ) { // O ( 10*2*log(k)) 
                UserTweetList ut = heap.poll(); // poll the list containing the latest tweet
                ret.add(ut.getVal()); // add the latest tweet to returning list
                // if the list has next one, put back to the heap
                if (ut.peek() != null) heap.offer(ut); // 
            return ret;
        /** Follower follows a followee. If the operation is invalid, it should be a no-op. */
        public void follow(int followerId, int followeeId) { // O(1)
            if (! follows.containsKey(followerId) ) {
                follows.put(followerId, new HashSet<Integer>());
        /** Follower unfollows a followee. If the operation is invalid, it should be a no-op. */
        public void unfollow(int followerId, int followeeId) { //O(1)
            if ( (followerId != followeeId) && (follows.containsKey(followerId)  ) ) {

  • 0

    I just tried it took 160ms but only beat 66%, sry bro :)
    I am wondering how people do it in 130ms

Log in to reply

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.