Design Hit Counter

Published: Sep 26, 2022

Problem Description

Design a hit counter which counts the number of hits received in the past 5 minutes (i.e., the past 300 seconds).

Your system should accept a timestamp parameter (in seconds granularity), and you may assume that calls are being made to the system in chronological order (i.e., timestamp is monotonically increasing). Several hits may arrive roughly at the same time.

Implement the HitCounter class:

• HitCounter() Initializes the object of the hit counter system.
• void hit(int timestamp) Records a hit that happened at timestamp (in seconds). Several hits may happen at the same timestamp.
• int getHits(int timestamp) Returns the number of hits in the past 5 minutes from timestamp (i.e., the past 300 seconds).

Constraints:

• 1 <= timestamp <= 2 * 10**9
• All the calls are being made to the system in chronological order (i.e., timestamp is monotonically increasing).
• At most 300 calls will be made to hit and getHits.

https://leetcode.com/problems/design-hit-counter/

Examples

Example 1
Input
["HitCounter", "hit", "hit", "hit", "getHits", "hit", "getHits", "getHits"]
[[], [1], [2], [3], [4], [300], [300], [301]]
Output
[null, null, null, null, 3, null, 4, 3]

Explanation
HitCounter hitCounter = new HitCounter();
hitCounter.hit(1);       // hit at timestamp 1.
hitCounter.hit(2);       // hit at timestamp 2.
hitCounter.hit(3);       // hit at timestamp 3.
hitCounter.getHits(4);   // get hits at timestamp 4, return 3.
hitCounter.hit(300);     // hit at timestamp 300.
hitCounter.getHits(300); // get hits at timestamp 300, return 4.
hitCounter.getHits(301); // get hits at timestamp 301, return 3.

How to Solve

The key to solve this problem is how to maintain hits’ timestamp. Since the timestamp is monotonically increasing – sorted already, the binary search to find an answer looks a good approach. However, it can be much simpler – keep only hits from 300 seconds before.

The solution here creates a queue for timestamps. Both, hit and getHits maintains the values in queue. If the values are less than or equals to given timestamp - 300, it deletes all hits before timestamp - 300. The method, gitHits, returns the length of the queue. That’s all.

Solution

class HitCounter {
private:
    queue<int> q;

public:
    HitCounter() {}

    void hit(int timestamp) {
        q.push(timestamp);
        while (q.front() <= timestamp - 300) {
            q.pop();
        }
    }

    int getHits(int timestamp) {
        while (!q.empty() && q.front() <= timestamp - 300) {
            q.pop();
        }
        return q.size();
    }
};

import java.util.ArrayDeque;
import java.util.Queue;

public class HitCounter {
    private Queue<Integer> q = new ArrayDeque<>();
    public HitCounter() {}

    public void hit(int timestamp) {
        q.add(timestamp);
        while (q.peek() <= timestamp - 300) {
            q.poll();
        }
    }

    public int getHits(int timestamp) {
        while (!q.isEmpty() && q.peek() <= timestamp - 300) {
            q.poll();
        }
        return q.size();
    }
}

var HitCounter = function() {
    this.queue = []
};

/**
 * @param {number} timestamp
 * @return {void}
 */
HitCounter.prototype.hit = function(timestamp) {
    this.queue.push(timestamp);
    while (this.queue[0] <= timestamp - 300) {
        this.queue.shift();
    }
};

/**
 * @param {number} timestamp
 * @return {number}
 */
HitCounter.prototype.getHits = function(timestamp) {
    while (this.queue[0] <= timestamp - 300) {
        this.queue.shift();
    }
    return this.queue.length;
};

class HitCounter:

    def __init__(self):
        self.queue = []

    def hit(self, timestamp: int) -> None:
        self.queue.append(timestamp)
        while self.queue[0] <= timestamp - 300:
            self.queue.pop(0)

    def getHits(self, timestamp: int) -> int:
        while self.queue and self.queue[0] <= timestamp - 300:
            self.queue.pop(0)
        return len(self.queue)

class HitCounter
  def initialize()
    @q = Array.new
  end


=begin
    :type timestamp: Integer
    :rtype: Void
=end
  def hit(timestamp)
    @q << timestamp
    while @q[0] <= timestamp - 300
      @q.shift
    end
  end


=begin
    :type timestamp: Integer
    :rtype: Integer
=end
  def get_hits(timestamp)
    while !@q.empty? && @q[0] <= timestamp - 300
      @q.shift
    end
    @q.size
  end


end

Complexities

• Time: O(n) – n: number of 300 or more older timestamps
• Space: O(m) – m: number of timestamps in a queue