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 timet, wheretrepresents some time in milliseconds, and returns the number of requests that has happened in the past3000milliseconds (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 <= 10<sup>9</sup>- Each test case will call
pingwith strictly increasing values oft. - At most
10<sup>4</sup>calls will be made toping.
/**
* 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();
}
}
