Saturday, June 11, 2016

LeetCode 355 - Design Twitter


https://leetcode.com/problems/design-twitter/
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. Your design should support the following methods:
  1. postTweet(userId, tweetId): Compose a new tweet.
  2. getNewsFeed(userId): 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.
  3. follow(followerId, followeeId): Follower follows a followee.
  4. unfollow(followerId, followeeId): Follower unfollows a followee.
Example:
Twitter twitter = new Twitter();

// User 1 posts a new tweet (id = 5).
twitter.postTweet(1, 5);

// User 1's news feed should return a list with 1 tweet id -> [5].
twitter.getNewsFeed(1);

// User 1 follows user 2.
twitter.follow(1, 2);

// User 2 posts a new tweet (id = 6).
twitter.postTweet(2, 6);

// 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.getNewsFeed(1);

// User 1 unfollows user 2.
twitter.unfollow(1, 2);

// User 1's news feed should return a list with 1 tweet id -> [5],
// since user 1 is no longer following user 2.
twitter.getNewsFeed(1);
http://bookshadow.com/weblog/2016/06/11/leetcode-design-twitter/
Map + Set + PriorityQueue
系统应当维护下列信息:
1). 一个全局的推文计数器:postCount 用户发推文时,计数器+1

2). 推文Id -> 推文计数器的映射:tweetIdMap 用来记录推文的次序

3). 推文Id -> 用户Id的映射:tweetOwnerMap 用来记录推文的发送者是谁

4). 用户Id -> 关注对象集合的映射: followeeMap 用来记录用户的关注对象(关注/取消关注)
方法的实现:
1). 关注 follow / 取消关注 unfollow:
只需要维护followeeMap中对应的关注对象集合followeeSet即可
2). 发送推文 postTweet:
更新推文计数器postCount,维护tweetIdMap;

向用户的推文发送列表中新增一条记录
3). 获取推文推送 getNewsFeed:
这里需要使用优先队列PriorityQueue,记为pq

遍历用户的关注对象列表,将每一位关注对象的最新一条推文添加至pq中。

然后从pq中弹出最近的一条推文,记为topTweetId;

通过tweetOwnerMap找到这条推文的发送者userId,然后将该发送者的下一条比较新的推文添加至pq。

直到弹出10条最新的推文,或者pq为空为止。
private int postCount; private Map<Integer, Integer> tweetCountMap; private Map<Integer, List<Integer>> tweetIdMap; private Map<Integer, Integer> tweetOwnerMap; private Map<Integer, Set<Integer>> followeeMap; /** Initialize your data structure here. */ public Twitter() { tweetCountMap = new HashMap<Integer, Integer>(); tweetIdMap = new HashMap<Integer, List<Integer>>(); tweetOwnerMap = new HashMap<Integer, Integer>(); followeeMap = new HashMap<Integer, Set<Integer>>(); } /** Compose a new tweet. */ public void postTweet(int userId, int tweetId) { tweetCountMap.put(tweetId, ++postCount); tweetOwnerMap.put(tweetId, userId); getTweetIdList(userId).add(tweetId); } /** 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> result = new ArrayList<Integer>(); Set<Integer> followeeSet = getFolloweeSet(userId); PriorityQueue<Integer> pq = new PriorityQueue<Integer>(new Comparator<Integer>(){ @Override public int compare(Integer a, Integer b) { return tweetCountMap.get(b) - tweetCountMap.get(a); } }); Map<Integer, Integer> idxMap = new HashMap<Integer, Integer>(); for (Integer followeeId : followeeSet) { List<Integer> tweetIdList = getTweetIdList(followeeId); int idx = tweetIdList.size() - 1; if (idx >= 0) { idxMap.put(followeeId, idx - 1); pq.add(tweetIdList.get(idx)); } } while (result.size() < 10 && !pq.isEmpty()) { Integer topTweetId = pq.poll(); result.add(topTweetId); Integer ownerId = tweetOwnerMap.get(topTweetId); int idx = idxMap.get(ownerId); if (idx >= 0) { List<Integer> tweetIdList = getTweetIdList(ownerId); pq.add(tweetIdList.get(idx)); idxMap.put(ownerId, idx - 1); } } return result; } /** Follower follows a followee. If the operation is invalid, it should be a no-op. */ public void follow(int followerId, int followeeId) { getFolloweeSet(followerId).add(followeeId); } /** Follower unfollows a followee. If the operation is invalid, it should be a no-op. */ public void unfollow(int followerId, int followeeId) { if (followerId != followeeId) { getFolloweeSet(followerId).remove(followeeId); } } /** Get a non-empty followee set of an user. */ private Set<Integer> getFolloweeSet(int userId) { Set<Integer> followeeSet = followeeMap.get(userId); if (followeeSet == null) { followeeSet = new HashSet<Integer>(); followeeSet.add(userId); followeeMap.put(userId, followeeSet); } return followeeSet; } /** Get a non-empty tweet id list of an user. */ private List<Integer> getTweetIdList(int userId) { List<Integer> tweetIdList = tweetIdMap.get(userId); if (tweetIdList == null) { tweetIdList = new ArrayList<Integer>(); tweetIdMap.put(userId, tweetIdList); } return tweetIdList; }
https://leetcode.com/discuss/107638/java-solution-using-hashmap-%26-hashset-%26-priorityqueue
private Map<Integer,Set<Integer>> users = new HashMap<>(); private Map<Integer,Map<Integer,Integer>> tweets = new HashMap<>(); private int timeStamp = 0; public Twitter() { } public void postTweet(int userId, int tweetId) { if(users.get(userId) == null){ users.put(userId,new HashSet<>()); tweets.put(userId,new HashMap<>()); } tweets.get(userId).put(timeStamp++,tweetId); } public List<Integer> getNewsFeed(int userId) { List<Integer> res = new ArrayList<>(); if(users.get(userId) == null) return res; Queue<Map.Entry<Integer,Integer>> queue = new PriorityQueue<>((e1,e2) -> e2.getKey() - e1.getKey()); for(Map.Entry<Integer,Integer> e : tweets.get(userId).entrySet()) queue.offer(e); for(Integer user : users.get(userId)){ for(Map.Entry<Integer,Integer> e : tweets.get(user).entrySet()){ queue.offer(e); } } for(int i = 0; i < 10 && !queue.isEmpty(); i++){ res.add(queue.poll().getValue()); } return res; } public void follow(int followerId, int followeeId) { if(followerId == followeeId) return; if(users.get(followerId) == null){ users.put(followerId,new HashSet<>()); tweets.put(followerId,new HashMap<>()); } if(users.get(followeeId) == null){ users.put(followeeId,new HashSet<>()); tweets.put(followeeId,new HashMap<>()); } users.get(followerId).add(followeeId); } public void unfollow(int followerId, int followeeId) { if(followerId == followeeId) return; if(users.get(followerId) == null || users.get(followeeId) == null) return; users.get(followerId).remove(followeeId); }
https://leetcode.com/discuss/107697/java-solution-using-hashmap-and-priorityqueue
private static class Tweet { int timestamp; int tweetId; public Tweet(int tweetId, int timestamp) { this.tweetId = tweetId; this.timestamp = timestamp; } } private Map<Integer, Set<Integer>> followMap = new HashMap<Integer, Set<Integer>>(); private Map<Integer, List<Tweet>> tweetMap = new HashMap<Integer, List<Tweet>>(); private AtomicInteger timestamp; /** Initialize your data structure here. */ public Twitter() { timestamp = new AtomicInteger(0); } /** Compose a new tweet. */ public void postTweet(int userId, int tweetId) { Tweet newTweet = new Tweet(tweetId, timestamp.getAndIncrement()); if (!tweetMap.containsKey(userId)) { tweetMap.put(userId, new ArrayList<Tweet>()); //Assuming no deletion for now? } tweetMap.get(userId).add(newTweet); } /** * 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> result = new ArrayList<Integer>(10); PriorityQueue<int[]> pq = new PriorityQueue<int[]>(new Comparator<int[]>() { public int compare(int[] it1, int[] it2) { return tweetMap.get(it2[0]).get(it2[1]).timestamp - tweetMap.get(it1[0]).get(it1[1]).timestamp; } }); Set<Integer> followeeSet = new HashSet<Integer>(); followeeSet.add(userId); if (followMap.containsKey(userId)) { followeeSet.addAll(followMap.get(userId)); } for (Integer followee : followeeSet) { if (tweetMap.containsKey(followee)) { List<Tweet> tweetList = tweetMap.get(followee); if (tweetList.size() > 0) { pq.add(new int[] { followee, tweetList.size() - 1 }); } } } while (result.size() < 10 && pq.size() > 0) { int[] it = pq.poll(); result.add(tweetMap.get(it[0]).get(it[1]).tweetId); it[1]--; if (it[1] >= 0) { pq.add(it); } } return result; } /** * Follower follows a followee. If the operation is invalid, it should be a * no-op. */ public void follow(int followerId, int followeeId) { Set<Integer> followSet = followMap.get(followerId); if (followSet == null) { followSet = new HashSet<Integer>(); followMap.put(followerId, followSet); } followSet.add(followeeId); } /** * Follower unfollows a followee. If the operation is invalid, it should be * a no-op. */ public void unfollow(int followerId, int followeeId) { Set<Integer> followSet = followMap.get(followerId); if (followSet == null) { followSet = new HashSet<Integer>(); followMap.put(followerId, followSet); } followSet.remove(followeeId); }
http://www.w2bc.com/article/149069
用户之间的follow关系因为userId是唯一的,所以可以用HashMap保存user之间的follow关系。
如何设计用户的feed流呢?看到本题的tag有heap,想了一下没想到如何用heap实现feed, 那就暴力点,将所有用户发布的feed 用一个list保存。 当要获取某用户的feeds时,按时间顺序从list后往前,如果一个feed的userid属于该用户或其follower则 将该feed存入结果集。
题目不难,甚至有点简单。。follow和unfollow时间复杂度为O(1),空间复杂度为O(n),getNewsFeed时间复杂度为O(n),空间复杂度为O(1)。
    HashMap<Integer, Set<Integer>> maps = new HashMap<Integer, Set<Integer>>();
    List<Feed> feeds = new ArrayList<Feed>();

    class Feed {
        int userId, tweetId;

        public Feed(int userId, int tweetId) {
            this.userId = userId;
            this.tweetId = tweetId;
        }

        public String toString(){
            return tweetId+"";
        }
    }

    /** Initialize your data structure here. */
    public Twitter() {

    }

    /** Compose a new tweet. */
    public void postTweet(int userId, int tweetId) {
        Feed f = new Feed(userId, tweetId);
        feeds.add(f);
    }

    /**
     * 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> res = new ArrayList<Integer>();
        Set<Integer> users = maps.get(userId);
        if(users==null)
            users=  new HashSet<Integer>();
        users.add(userId);
        for (int i=feeds.size()-1;i>=0;i--) {
            Feed f = feeds.get(i);
            if (res.size()<10&&users.contains(f.userId)) {
                res.add(f.tweetId);
            }
            if(res.size()>=10)
                break;
        }
        return res;
    }

    /**
     * Follower follows a followee. If the operation is invalid, it should be a
     * no-op.
     */
    public void follow(int followerId, int followeeId) {
        Set<Integer> sets;
        if (maps.containsKey(followerId)) {
            sets = maps.get(followerId);
        } else {
            sets = new HashSet<Integer>();
        }
        sets.add(followeeId);
        maps.put(followerId, sets);
    }

    /**
     * Follower unfollows a followee. If the operation is invalid, it should be
     * a no-op.
     */
    public void unfollow(int followerId, int followeeId) {
        Set<Integer> sets = maps.get(followerId);
        if(sets!=null&&sets.contains(followeeId)){
            sets.remove(followeeId);
            maps.put(followerId, sets);
        }
    }
https://www.hrwhisper.me/leetcode-design-twitter/
题意:要求设计一个数据结构,使其能满足twitter的4种基本操作,发推、获得关注用户和自身最新10条推文、关注用户和取消关注。
思路:水题。和上一题一样,不明白为啥为hard。  可能难点就在于,直接写出无bug 的code吧。
需要那些非法的情况需要进行考虑吧。比如:
  • follow操作 followerId, followeeId 相等
  • unfollow操作followerId 不存在,或者followerId 压根就没关注 followeeId
求10个最近的推文可以用堆(当然这里我没有)

Python
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class Twitter(object):
    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.tweets_cnt = 0
        self.tweets = collections.defaultdict(list)
        self.follower_ship = collections.defaultdict(set)
    def postTweet(self, userId, tweetId):
        """
        Compose a new tweet.
        :type userId: int
        :type tweetId: int
        :rtype: void
        """
        self.tweets[userId].append([tweetId, self.tweets_cnt])
        self.tweets_cnt += 1
    def getNewsFeed(self, userId):
        """
        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.
        :type userId: int
        :rtype: List[int]
        """
        recent_tweets = []
        user_list = list(self.follower_ship[userId]) + [userId]
        userId_tweet_index = [[userId, len(self.tweets[userId]) - 1] for userId in user_list if userId in self.tweets]
        for _ in xrange(10):
            max_index = max_tweet_id = max_user_id = -1
            for i, (user_id, tweet_index) in enumerate(userId_tweet_index):
                if tweet_index >= 0:
                    tweet_info = self.tweets[user_id][tweet_index]
                    if tweet_info[1] > max_tweet_id:
                        max_index, max_tweet_id, max_user_id = i, tweet_info[1], user_id
            if max_index < 0: break
            recent_tweets.append(self.tweets[max_user_id][userId_tweet_index[max_index][1]][0])
            userId_tweet_index[max_index][1] -= 1
        return recent_tweets
    def follow(self, followerId, followeeId):
        """
        Follower follows a followee. If the operation is invalid, it should be a no-op.
        :type followerId: int
        :type followeeId: int
        :rtype: void
        """
        if followerId != followeeId:
            self.follower_ship[followerId].add(followeeId)
    def unfollow(self, followerId, followeeId):
        """
        Follower unfollows a followee. If the operation is invalid, it should be a no-op.
        :type followerId: int
        :type followeeId: int
        :rtype: void
        """
        if followerId in self.follower_ship and followeeId in self.follower_ship[followerId]:
            self.follower_ship[followerId].remove(followeeId)






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How To (1) Algorithm - New (1) Algorithm Series (1) Algorithms Part I (1) Analysis of Algorithm (1) Array-Element Index Negative (1) Array-Rearrange (1) Auxiliary Array (1) Auxiliary Array: Inc&Dec (1) BACK (1) BK-Tree (1) BZOJ (1) Basic (1) Bayes (1) Beauty of Math (1) Big Integer (1) Big Number (1) Binary (1) Binary Tree Variant (1) Bipartite (1) Bit-Missing Number (1) BitMap (1) BitMap index (1) BitSet (1) Bug Free Code (1) BuildIt (1) C/C++ (1) CC Interview (1) Cache (1) Calculate Height at Same Recusrion (1) Cartesian tree (1) Check Tree Property (1) Chinese (1) Circular Buffer (1) Code Quality (1) Codesolutiony (1) Company - Alibaba (1) Company - Palantir (1) Company - WalmartLabs (1) Company-Apple (1) Company-Epic (1) Company-Salesforce (1) Company-Snapchat (1) Company-Yelp (1) Compression Algorithm (1) Concurrency (1) Convert BST to DLL (1) Convert DLL to BST (1) Custom Sort (1) Cyclic Replacement (1) DFS-Matrix (1) DP - Probability (1) DP Fill Diagonal First (1) DP-Difficult (1) DP-End with 0 or 1 (1) DP-Fill Diagonal First (1) DP-Graph (1) DP-Left and Right Array (1) DP-MaxMin (1) DP-Memoization (1) DP-Node All Possibilities (1) DP-Optimization (1) DP-Preserve Previous Value (1) DP-Print All Solution (1) Database (1) Detect Negative Cycle (1) Directed Graph (1) Do Two Things at Same Recusrion (1) Domino (1) Dr Dobb's (1) Duplicate (1) Equal probability (1) External Sort (1) FST (1) Failure Function (1) Fraction (1) Front End Pointers (1) Funny (1) Fuzzy String Search (1) Game (1) Generating Function (1) Generation (1) Genetic algorithm (1) GeoHash (1) Geometry - Orientation (1) Google APAC (1) Graph But No Graph (1) Graph Transpose (1) Graph Traversal (1) Graph-Coloring (1) Graph-Longest Path (1) Gray Code (1) HOJ (1) Hanoi (1) Hard Algorithm (1) How Hash (1) How to Test (1) Improve It (1) In Place (1) Inorder-Reverse Inorder Traverse Simultaneously (1) Interpolation search (1) Interview (1) Interview - Easy (1) Interview - Facebook (1) Isomorphic (1) JDK8 (1) K Dimensional Tree (1) Knapsack - Fractional (1) Knapsack - ZeroOnePack (1) Knight (1) Kosaraju’s algorithm (1) Kruskal (1) Kruskal MST (1) Kth Element (1) Least Common Ancestor (1) LeetCode - Binary Tree (1) LeetCode - Coding (1) LeetCode - Detail (1) LeetCode - Related (1) LeetCode Diffcult (1) Linked List Reverse (1) Linkedin (1) Linkedin Interview (1) Local MinMax (1) Logic Pattern (1) Longest Common Subsequence (1) Longest Common Substring (1) Longest Prefix Suffix(LPS) (1) Manhattan Distance (1) Map && Reverse Map (1) Math - Induction (1) Math-Multiply (1) Math-Sum Of Digits (1) Matrix - O(N+M) (1) Matrix BFS (1) Matrix Graph (1) Matrix Search (1) Matrix+DP (1) Matrix-Rotate (1) Max Min So Far (1) Median (1) Memory-Efficient (1) MinHash (1) MinMax Heap (1) Monotone Queue (1) Monto Carlo (1) Multi-Reverse (1) Multiple DFS (1) Multiple Tasks (1) Next Successor (1) Offline Algorithm (1) PAT (1) Partition (1) Path Finding (1) Patience Sort (1) Persistent (1) Pigeon Hole Principle (1) Power Set (1) Pratical Algorithm (1) Probabilistic Data Structure (1) Proof (1) Python (1) Queue & Stack (1) RSA (1) Ranking (1) Rddles (1) ReHash (1) Realtime (1) Recurrence Relation (1) Recursive DFS (1) Recursive to Iterative (1) Red-Black Tree (1) Region (1) Regular Expression (1) Resources (1) Reverse Inorder Traversal (1) Robin (1) Selection (1) Self Balancing BST (1) Similarity (1) Sort && Binary Search (1) String Algorithm. Symbol Table (1) String DP (1) String Distance (1) SubMatrix (1) Subsequence (1) System of Difference Constraints(差分约束系统) (1) TSP (1) Ternary Search Tree (1) Test (1) Thread (1) TimSort (1) Top-Down (1) Tournament (1) Tournament Tree (1) Transform Tree in Place (1) Tree Diameter (1) Tree Rotate (1) Trie + DFS (1) Trie and Heap (1) Trie vs Hash (1) Trie vs HashMap (1) Triplet (1) Two Data Structures (1) Two Stacks (1) USACO - Classical (1) USACO - Problems (1) UyHiP (1) Valid Tree (1) Vector (1) Wiggle Sort (1) Wikipedia (1) Yahoo Interview (1) ZOJ (1) baozitraining (1) codevs (1) cos126 (1) javabeat (1) jum (1) namic Programming (1) sqrt(N) (1) 两次dijkstra (1) 九度 (1) 二进制枚举 (1) 夹逼法 (1) 归一化 (1) 折半枚举 (1) 枚举 (1) 状态压缩DP (1) 男人八题 (1) 英雄会 (1) 逆向思维 (1)

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