933. Number of Recent Calls

Description

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

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.

Submitted Code

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)
            
            
# Your RecentCounter object will be instantiated and called as such:
# obj = RecentCounter()
# param_1 = obj.ping(t)

Runtime: 22 ms | Beats 95.31%
Memory: 23.13 MB | Beats 41.34%

ping에 매개변수로 들어오는 값이 점점 높아지는 조건이기 때문에 현재 범위의 최소값보다 작은 request는 제외해야 효율이 높아진다. 리스트로도 구현할 수 있지만 더 빠른 연산을 위해 collections 모듈의 deque()를 이용했다.

Other Solutions

1st

class RecentCounter:

    def __init__(self):
        self.counter =0
        self.queue =deque()

    def ping(self, t: int) -> int:
        self.queue.append(t)
        self.counter+=1

        while self.queue[0] < t-3000:
            self.queue.popleft()
            self.counter-=1

        return self.counter

time complexity: 𝑂(1) ← per operation
space complexity: 𝑂(𝑛)

ping()을 호출할 때마다 큐의 길이를 세는 것보다 더 효율적인 방법인 것 같다.