933. Number of Recent Calls

Problem statement:

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.

Source: LeetCode

Solution:

First solution: Using queue

if we think properly it’s a problem that can be solved by queue because we need the number of requests made in the last 3000 seconds, so we can keep polling the queue till the time we have values satisfy condition t < t’ – 3000

Java Code

class RecentCounter {

    Queue<Integer> q = new LinkedList<>();

    public RecentCounter() {
        q = new LinkedList<>();
    }

    public int ping(int t) {
        q.add(t);
        while (q.peek() < t - 3000) {
            q.poll();
        }
        return q.size();
    }
}

• Space complexity: O(N)
• Time complexity: O(Min (N, 3000))

Second solution: Using TreeSet

Now as we see in the normal queue we were searching elements in linear search. But using SortedSet (TreeSet) we can add an element and search subset in O(log(N))

Java Code

class RecentCounter {

TreeSet<Integer> treeSet;
public RecentCounter() {
    treeSet = new TreeSet();
}

public int ping(int t) {
    treeSet.add(t);
    return treeSet.tailSet(t-3000).size();
}
}

• Space complexity: O(N)
• Time complexity: O(log (N))