3662.design-hit-counter.py

#  Tag: Simulation
#  Time: O(N)
#  Space: O(N)
#  Ref: Leetcode-362
#  Note: -

#  In this question, you need to design a HitCounter class.
#  
#  In this class, there are the following functions:
#  
#  - `HitCounter()`: No-argument constructor
#  - `void hit(int timestamp)`: Indicates that a tap occurs at the specified time
#  - `int getHits(int timestamp)`: Returns the total number of hits within 300 seconds before the specified time
#  
#  Where `timestamp` is in seconds.
#  
#  **Example 1**
#  
#  Input:
#  
#  ```plaintext
#  ["hit(1)", "hit(2)", "hit(3)", "getHits(3)", "hit(301)", "getHits(301)"]
#  ```
#  
#  Output:
#  
#  ```plaintext
#  [null, null, null, 3, null, 3]
#  ```
#  
#  **Example 2**
#  
#  Input:
#  
#  ```plaintext
#  ["hit(1)", "hit(1)", "hit(1)", "getHits(2)"]
#  ```
#  
#  Output:
#  
#  ```plaintext
#  [null, null, null, 3]
#  ```
#  
#  You can assume that when the system calls the `void hit(int timestamp)` and `int getHits(int timestamp)` methods, it must be called in chronological order, that is, the smaller `timestamp` is called first.
#  
#  There may be multiple hits at the same time.
#  
#  Hit action takes negligible time.
#  
#  At the same moment (`timestamp`), `getHits()` will not happen before `hit()`.

from collections import deque
class HitCounter:

    def __init__(self):
        self.q = deque()

    """
    @param timestamp: the timestamp
    @return: nothing
    """
    def hit(self, timestamp: int):
        self.q.append(timestamp)

    """
    @param timestamp: the timestamp
    @return: the count of hits in recent 300 seconds
    """
    def get_hits(self, timestamp: int) -> int:
        while len(self.q) > 0 and self.q[0] <= timestamp - 300:
            self.q.popleft()

        return len(self.q)