Monday, August 8, 2016

LeetCode 362 - Design Hit Counter



Design a hit counter which counts the number of hits received in the past 5 minutes.
Each function accepts a timestamp parameter (in seconds granularity) and you may assume that calls are being made to the system in chronological order (ie, the timestamp is monotonically increasing). You may assume that the earliest timestamp starts at 1.
It is possible that several hits arrive roughly at the same time.
Example:
HitCounter counter = new HitCounter();

// hit at timestamp 1.
counter.hit(1);

// hit at timestamp 2.
counter.hit(2);

// hit at timestamp 3.
counter.hit(3);

// get hits at timestamp 4, should return 3.
counter.getHits(4);

// hit at timestamp 300.
counter.hit(300);

// get hits at timestamp 300, should return 4.
counter.getHits(300);

// get hits at timestamp 301, should return 3.
counter.getHits(301);
http://www.cnblogs.com/grandyang/p/5605552.html
这道题让我们设计一个点击计数器,能够返回五分钟内的点击数,提示了有可能同一时间内有多次点击。由于操作都是按时间顺序的,下一次的时间戳都会大于等于本次的时间戳,那么最直接的方法就是用一个队列queue,每次点击时都将当前时间戳加入queue中,然后在需要获取点击数时,我们从队列开头开始看,如果开头的时间戳在5分钟以外了,就删掉,直到开头的时间戳在5分钟以内停止,然后返回queue的元素个数即为所求的点击数

    void hit(int timestamp) {
        q.push(timestamp);
    }
    
    /** Return the number of hits in the past 5 minutes.
        @param timestamp - The current timestamp (in seconds granularity). */
    int getHits(int timestamp) {
        while (!q.empty() && timestamp - q.front() >= 300) {
            q.pop();
        }
        return q.size();
    }

private:
    queue<int> q;

X. https://discuss.leetcode.com/topic/48758/super-easy-design-o-1-hit-o-s-gethits-no-fancy-data-structure-is-needed
O(s) s is total seconds in given time interval, in this case 300.
basic ideal is using buckets. 1 bucket for every second because we only need to keep the recent hits info for 300 seconds. hit[] array is wrapped around by mod operation. Each hit bucket is associated with times[] bucket which record current time. If it is not current time, it means it is 300s or 600s... ago and need to reset to 1.
public class HitCounter {
    private int[] times;
    private int[] hits;
    /** Initialize your data structure here. */
    public HitCounter() {
        times = new int[300];
        hits = new int[300];
    }
    
    /** Record a hit.
        @param timestamp - The current timestamp (in seconds granularity). */
    public void hit(int timestamp) {
        int index = timestamp % 300;
        if (times[index] != timestamp) {
            times[index] = timestamp;
            hits[index] = 1;
        } else {
            hits[index]++;
        }
    }
    
    /** Return the number of hits in the past 5 minutes.
        @param timestamp - The current timestamp (in seconds granularity). */
    public int getHits(int timestamp) {
        int total = 0;
        for (int i = 0; i < 300; i++) {
            if (timestamp - times[i] < 300) {
                total += hits[i];
            }
        }
        return total;
    }
}

由于Follow up中说每秒中会有很多点击,下面这种方法就比较巧妙了,定义了两个大小为300的一维数组times和hits,分别用来保存时间戳和点击数,在点击函数中,将时间戳对300取余,然后看此位置中之前保存的时间戳和当前的时间戳是否一样,一样说明是同一个时间戳,那么对应的点击数自增1,如果不一样,说明已经过了五分钟了,那么将对应的点击数重置为1。那么在返回点击数时,我们需要遍历times数组,找出所有在5分中内的位置,然后把hits中对应位置的点击数都加起来即可

final int  fiveMin = 300;
LinkedHashMap<Integer,Integer> map;
/** Initialize your data structure here. */
public HitCounter() {
    map = new LinkedHashMap<Integer,Integer>(){
        @Override
        protected boolean removeEldestEntry(Map.Entry<Integer,Integer> eldest){
            return map.size() > fiveMin;
        }
        };
}

/** Record a hit.
    @param timestamp - The current timestamp (in seconds granularity). */
public void hit(int timestamp) {
    map.put(timestamp,map.getOrDefault(timestamp,0)+1);
}

/** Return the number of hits in the past 5 minutes.
    @param timestamp - The current timestamp (in seconds granularity). */
public int getHits(int timestamp) {
    int start = timestamp - fiveMin;
    int sum =0;
    for(int tsp:map.keySet()){
        if(tsp>start)
            sum += map.get(tsp);
    }
    return sum;
}
public class HitCounter 
{
    int winLo = 0, winHi = 0; 
    LinkedList<int[]> win = new LinkedList<int[]>();

    public HitCounter() {
    }
    
    public void hit(int timestamp) 
    {
        winHi++;
        
        if(win.size() == 0 || win.getLast()[0] != timestamp)
            win.add(new int[] { timestamp, winHi});
        else
            win.getLast()[1] = winHi;
            
        ShrinkWin(timestamp);
    }
    
    public int getHits(int timestamp) 
    {
        ShrinkWin(timestamp);
        return (int)(winHi - winLo);
    }
    
    private void ShrinkWin(int timestamp)
    {
        while (win.size() > 0 && timestamp - win.getFirst()[0] >= 300) 
        {
            winLo = win.getFirst()[1];
            win.removeFirst();
        }        
    }
}
X. https://discuss.leetcode.com/topic/48752/simple-java-solution-with-explanation
-- use too much space
In this problem, I use a queue to record the information of all the hits. Each time we call the function getHits( ), we have to delete the elements which hits beyond 5 mins (300). The result would be the length of the queue : )
public class HitCounter {
        Queue<Integer> q = null;
        /** Initialize your data structure here. */
        public HitCounter() {
            q = new LinkedList<Integer>();
        }
        
        /** Record a hit.
            @param timestamp - The current timestamp (in seconds granularity). */
        public void hit(int timestamp) {
            q.offer(timestamp);
        }
        
        /** Return the number of hits in the past 5 minutes.
            @param timestamp - The current timestamp (in seconds granularity). */
        public int getHits(int timestamp) {
            while(!q.isEmpty() && timestamp - q.peek() >= 300) {
                q.poll();
            }
            return q.size();
        }
    }

X. Extended
https://reeestart.wordpress.com/2016/06/12/google-last-request-number/
Interview中的一道常见题。就是要求设计数据结构来返回lastSecond(), lastMinute(), lastHour()的request数量。
因为之前太想当然的假设在多线程环境下执行,完全忽略了在sequential环境下很多的edge case.
Circular Buffer的确是这道题的考点,但是如果call getHits()的时间距离上一次hit()的时间早就超过了5分钟,那circular buffer里的旧数据要怎么更新是个大问题。
一开始想着记录最近一次hit的时间,但是不够。如果当前时间距离lastHit超过5分钟了,那好办,整个circular buffer清零就好。但是如果没有超过5min,circular buffer依然要update,这样的话只有lastHit是不够的,因为没有办法找到距离当前时间5min以前的hits,也就根本没法知道circular buffer里哪些需要update,哪些不需要。
这道题只有一个circular buffer是不够的。
1. 要么用Deque做并wrap一个class包含时间和hit count (不能用queue是因为要获取最近一次hit的时间,而queue应该只能peek head)
2. 要么在circular buffer的基础上再加一个time[] 数组记录circular buffer里每个cell的时间。
[Time & Space]
两种方法的query time都为O(n),hit()为O(1).
space也都为O(n)
// circular buffer
class HitCounter {
 
  int[] buffer;
  int[] time;
 
  public HitCounter() {
    this.buffer = new int[5 * 60];
    this.time = new int[5 * 60];
  }
 
  public void hit(int timestamp) {
    int idx = timestamp % 300;
    if (timestamp == time[idx]) {
      buffer[idx]++;
    }
    else {
      buffer[idx] = 1;
      time[idx] = timestamp;
    }
  }
 
  public int getHits(int timestamp) {
    int result = 0;
    for (int i = 0; i < 300; i++) {
      if (timestamp - time[i] < 300) {
        result += buffer[i];
      }
    }
 
    return result;
  }
}
 
// Queue + Wrapper class
class HitCounter2 {
 
  Deque<Hit> deque;
  int hits = 0;
 
  public HitCounter2() {
    this.deque = new LinkedList<>();
  }
 
  public void hit(int timestamp) {
    if (!deque.isEmpty() && timestamp == deque.getLast().time) {
      deque.getLast().cnt++;
    }
    else {
      deque.offerLast(new Hit(timestamp));
    }
    hits++;
  }
 
  public int getHits(int timestamp) {
    while (!deque.isEmpty() && timestamp - deque.getFirst().time >= 300) {
      hits -= deque.pollFirst().cnt;
    }
 
    return hits;
  }
 
  private class Hit {
    int time;
    int cnt;
 
    public Hit(int time) {
      this.time = time;
      this.cnt = 1;
    }
  }
}


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