# 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
}

public int ping(int t) {

// add the item to the Queue

// 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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