Number of Recent Calls

QueueDesign

EasyLeetCode #933

15 min

Problem Statement

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

Example 1:

Operation Sequence:

ping(1)returns: 1

ping(100)returns: 2

ping(3001)returns: 3

ping(3002)returns: 3

Solution

A queue is the perfect data structure for this problem because we are interested in a sliding window of time. As new pings arrive, we can efficiently remove old pings that fall outside the 3000ms window from the front of the queue.

Algorithm Steps

Initialize a queue to store the timestamps of the pings.
When a new ping arrives at time `t`, add it to the queue.
Remove all timestamps from the front of the queue that are older than `t - 3000`.
The size of the queue is the number of recent calls.

Current Ping: N/A

Queue:

Start with an empty queue.

Number of Recent Calls Solution


from collections import deque

class RecentCounter:
    def __init__(self):
        self.requests = deque()

    def ping(self, t: int) -> int:
        self.requests.append(t)
        while self.requests[0] < t - 3000:
            self.requests.popleft()
        return len(self.requests)