# 933. Number of Recent Calls (Easy)

https://leetcode.com/problems/number-of-recent-calls/

Write a class `RecentCounter` to count recent requests.

It has only one method: `ping(int t)`, where t represents some time in milliseconds.

Return the number of `ping`s that have been made from 3000 milliseconds ago until now.

Any ping with time in `[t - 3000, t]` will count, including the current ping.

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

Example 1:

```Input: inputs = ["RecentCounter","ping","ping","ping","ping"], inputs = [[],[1],[100],[3001],[3002]]
Output: [null,1,2,3,3]```

Note:

1. Each test case will have at most `10000` calls to `ping`.
2. Each test case will call `ping` with strictly increasing values of `t`.
3. Each call to ping will have `1 <= t <= 10^9`.

## Solutions

``````class RecentCounter {
private Queue<Integer> queue;

public RecentCounter() {
// We can use PriorityQueue if incoming pings are out of order
// Since problem already clarified the every ping is larger than previous,
// no need to sort them and computed overhead.
}

public int ping(int t) {

// poll the expired ping out of the queue and conserve the space
while (queue.peek() < t - 3000) {
queue.poll();
}

return queue.size();
}
}
``````