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# Number of Recent Calls

## [Number of Recent Calls](https://leetcode.com/problems/number-of-recent-calls)

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`.

## Solutions

### 🧠 Cpp

```cpp
class RecentCounter
{
    list<int> pings;

public:
    RecentCounter() = default;

    int ping(int t)
    {

        for(auto iter = pings.begin(); iter != pings.end();)
        {
            if(t-*iter > 3000)
                iter = pings.erase(iter);
            else
                iter++;
        }
        pings.push_front(t);

        return pings.size();

    }
};

/**
 * Your RecentCounter object will be instantiated and called as such:
 * RecentCounter* obj = new RecentCounter();
 * int param_1 = obj->ping(t);
 */
```