Number of Recent Calls

This is a nice problem from LeetCode (Problem #933).

Problem: 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();
recentCounter.ping(1);     // requests = [1], range is [-2999,1], return 1
recentCounter.ping(100);   // requests = [1, 100], range is [-2900,100], return 2
recentCounter.ping(3001);  // requests = [1, 100, 3001], range is [1,3001], return 3
recentCounter.ping(3002);  // requests = [1, 100, 3001, 3002], range is [2,3002], return 3

Solution

We are interested in counting the number of requests in the inclusive range [$t - 3000, t$]. Therefore, we maintain the requests in a queue and we repeatedly pop off the first element if its value is $< t - 3000$, where $t$ is the current time. The remaining elements in the queue constitute the requests that happened in the inclusive range [$t - 3000, t$].

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

    def ping(self, t: int) -> int:
        self.counters.append(t) 
        while self.counters and self.counters[0] < t - 3000:  
            self.counters.popleft() 
        return len(self.counters) 
        
# Your RecentCounter object will be instantiated and called as such:
# obj = RecentCounter()
# param_1 = obj.ping(t)

Runtime:

The time complexity is $O(1)$. This is because a ping() operation involves just an append() to the queue and a popleft(), both of which are constant operations.

The space complexity is also $O(1)$ as our queue will never exceed 3000ms due to the problem definition.

Back to Queues Problems

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