Number of Recent Calls

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.

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.

Solution

The problem Number of Recent Calls can be solved by storing the time of each calls in a queue and popping every records that happened before the past 3000 milliseconds of a given call.

Implementation

static const int fast_io = []()
{
    std::ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    return 0;
}();

class RecentCounter
{
  private:
    queue<int> calls;

  public:
    RecentCounter() {}

    int ping(int t)
    {
        calls.push(t);
        while (calls.front() < t - 3000)
            calls.pop();

        return calls.size();
    }
};

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