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Number Of Recent Calls

Problem Statement​

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 have 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:​

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‘`1 <= t <= 10^9`
  • Eachtestcasewillcall‘ping‘withstrictlyincreasingvaluesof‘t‘.Each test case will call `ping` with strictly increasing values of `t`.
  • Atmost104callswillbemadeto‘ping‘.At most 10^4 calls will be made to `ping`.

Algorithm​

  1. Initialization: Each implementation initializes a list (records in Python, std::vector in C++, ArrayList in Java, and array in JavaScript) to store timestamps and a start index to track the beginning of the valid range.

  2. ping Method:

    • Appends the current timestamp to records.
    • Checks and updates start to maintain the invariant that timestamps within the last 3000 milliseconds are within the range [start, end].
    • Returns the number of valid timestamps (from start to end of records).

Python Implementation​

class RecentCounter:
def __init__(self):
self.records = []
self.start = 0

def ping(self, t: int) -> int:
self.records.append(t)
while self.records[self.start] < t - 3000:
self.start += 1
return len(self.records) - self.start

C++ Implementation​

#include <vector>

class RecentCounter {
private:
std::vector<int> records;
int start;
public:
RecentCounter() {
start = 0;
}

int ping(int t) {
records.push_back(t);
while (records[start] < t - 3000) {
start++;
}
return records.size() - start;
}
};

Java Implementation​

import java.util.ArrayList;
import java.util.List;

class RecentCounter {
private List<Integer> records;
private int start;

public RecentCounter() {
records = new ArrayList<>();
start = 0;
}

public int ping(int t) {
records.add(t);
while (records.get(start) < t - 3000) {
start++;
}
return records.size() - start;
}
}

JavaScript Implementation​

class RecentCounter {
constructor() {
this.records = [];
this.start = 0;
}

ping(t) {
this.records.push(t);
while (this.records[this.start] < t - 3000) {
this.start++;
}
return this.records.length - this.start;
}
}