RecentCounts.java

class RecentCounter {

    List<Integer> queue;
    int head;
    int tail;

    public RecentCounter() {
        this.queue = new ArrayList<Integer>();
        this.head = -1;
        this.tail = -1;
    }
    
    public int ping(int t) {
        add(t);
        remove(t);
        return (tail - head) + 1;
    }

    private void add(int t){
        if (queue.size() == 0){
            head++;
        }
        queue.add(t);
        tail++;
    }

    private void remove(int t){
        var low = t - 3000;
        while (
            (head <= tail) &&
            (low >= 0) &&
            (queue.get(head) < low)
        ){
            head++;
        }
    }
}

Leetcode

Link: https://leetcode.com/problems/number-of-recent-calls/description/?envType=study-plan-v2&envId=leetcode-75

Achievement

• The above uses queue logic to solve the problem.
• All the other Solutions found in the solutions section have the same runtime.
• The above solution beats 70% of the total submissions with a runtime value of 21 ms.