Solve the Number of Recent Calls Challenge in Java

The challenge

You have a RecentCounter class which counts the number of recent requests within a certain time frame.

Implement the RecentCounter class:

  • RecentCounter() Initializes the counter with zero recent requests.
  • int ping(int t) Adds a new request at time t, where t represents some time in milliseconds, and returns the number of requests that has happened in the past 3000 milliseconds (including the new request). Specifically, return the number of requests that have happened in the inclusive range [t - 3000, t].

It is guaranteed that every call to ping uses a strictly larger value of t than the previous call.

Example 1:

Input
  ["RecentCounter", "ping", "ping", "ping", "ping"]
  [[], [1], [100], [3001], [3002]]

Output
  [null, 1, 2, 3, 3]

Explanation
  RecentCounter recentCounter = new RecentCounter();

  // requests = [1], range is [-2999,1], return 1
  recentCounter.ping(1);
  
  // requests = [1, 100], range is [-2900,100], return 2
  recentCounter.ping(100);

  // requests = [1, 100, 3001], range is [1,3001], return 3
  recentCounter.ping(3001);

  // requests = [1, 100, 3001, 3002], range is [2,3002], return 3
  recentCounter.ping(3002);

Constraints:

  • 1 <= t <= 109
  • Each test case will call ping with strictly increasing values of t.
  • At most 104 calls will be made to ping.
/**
 * Your RecentCounter object will be instantiated and called as such:
 * RecentCounter obj = new RecentCounter();
 * int param_1 = obj.ping(t);
 */

The solution in Java

This problem requires using a Queue which we create from a LinkedList.

class RecentCounter {

    // create a Queue
    private Queue<Integer> q;
    
    public RecentCounter() {
        // initialise the Queue using a LinkedList
        q = new LinkedList<>();
    }
    
    public int ping(int t) {
        
        // add the item to the Queue
        q.add(t);
        
        // check if the front item is less than the item minus 3000 (our max)
        while(q.peek() < t - 3000) {
            // remove the item from the Queue
            q.poll();
        }
        
        // report the size back
        return q.size();
        
    }
}

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