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:
- Each test case will have at most
10000
calls toping
. - Each test case will call
ping
with strictly increasing values oft
. - 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.
queue = new LinkedList<>();
}
public int ping(int t) {
queue.add(t);
// poll the expired ping out of the queue and conserve the space
while (queue.peek() < t - 3000) {
queue.poll();
}
return queue.size();
}
}