# 362. Design Hit Counter / Medium

Design a hit counter which counts the number of hits received in the past 5 minutes (i.e., the past 300 seconds).

Your system should accept a timestamp parameter (in seconds granularity), and you may assume that calls are being made to the system in chronological order (i.e., timestamp is monotonically increasing). Several hits may arrive roughly at the same time.

Implement the HitCounter class:

• HitCounter() Initializes the object of the hit counter system.
• void hit(int timestamp) Records a hit that happened at timestamp (in seconds). Several hits may happen at the same timestamp.
• int getHits(int timestamp) Returns the number of hits in the past 5 minutes from timestamp (i.e., the past 300 seconds).

## Example 1:

Input
[“HitCounter”, “hit”, “hit”, “hit”, “getHits”, “hit”, “getHits”, “getHits”]
[[], [1], [2], [3], [4], [300], [300], [301]]
Output
[null, null, null, null, 3, null, 4, 3]

Explanation
HitCounter hitCounter = new HitCounter();
hitCounter.hit(1); // hit at timestamp 1.
hitCounter.hit(2); // hit at timestamp 2.
hitCounter.hit(3); // hit at timestamp 3.
hitCounter.getHits(4); // get hits at timestamp 4, return 3.
hitCounter.hit(300); // hit at timestamp 300.
hitCounter.getHits(300); // get hits at timestamp 300, return 4.
hitCounter.getHits(301); // get hits at timestamp 301, return 3.

## Constraints:

• 1 <= timestamp <= 2 * 10^9
• All the calls are being made to the system in chronological order (i.e., timestamp is monotonically increasing).
• At most 300 calls will be made to hit and getHits.

# Solution: Queue

## 思路

getHits 時，我們計算 queue 結尾的 timestamp 與開頭的 timestamp 差異是否超過 300，如果是，就不斷 pop 直到 queue 中只有這 300 秒的 hits。

## 效能

### Complexity

• Time Complexity:
• hit() - O(1)
• getHits() - O(1), worst O(N)
• Space Complexity: O(1)

### LeetCode Result

• Runtime: 0 ms
• Memory Usage: 7.4 MB