Find Servers That Handled Most Number of Requests

Problem

You have k servers numbered from 0 to k-1 that are being used to handle multiple requests simultaneously. Each server has infinite computational capacity but cannot handle more than one request at a time. The requests are assigned to servers according to a specific algorithm:

The ith (0-indexed) request arrives.
If all servers are busy, the request is dropped (not handled at all).
If the (i % k)th server is available, assign the request to that server.
Otherwise, assign the request to the next available server (wrapping around the list of servers and starting from 0 if necessary). For example, if the ith server is busy, try to assign the request to the (i+1)th server, then the (i+2)th server, and so on.

You are given a strictly increasing array arrival of positive integers, where arrival[i] represents the arrival time of the ith request, and another array load, where load[i] represents the load of the ith request (the time it takes to complete). Your goal is to find the busiest server(s). A server is considered busiest if it handled the most number of requests successfully among all the servers.

Return **a list containing the IDs (0-indexed) of the *busiest server(s)***. You may return the IDs in any order.

Example 1:

Input: k = 3, arrival = [1,2,3,4,5], load = [5,2,3,3,3] 
Output: [1] 
Explanation: 
All of the servers start out available.
The first 3 requests are handled by the first 3 servers in order.
Request 3 comes in. Server 0 is busy, so it's assigned to the next available server, which is 1.
Request 4 comes in. It cannot be handled since all servers are busy, so it is dropped.
Servers 0 and 2 handled one request each, while server 1 handled two requests. Hence server 1 is the busiest server.

Example 2:

Input: k = 3, arrival = [1,2,3,4], load = [1,2,1,2]
Output: [0]
Explanation: 
The first 3 requests are handled by first 3 servers.
Request 3 comes in. It is handled by server 0 since the server is available.
Server 0 handled two requests, while servers 1 and 2 handled one request each. Hence server 0 is the busiest server.

Example 3:

Input: k = 3, arrival = [1,2,3], load = [10,12,11]
Output: [0,1,2]
Explanation: Each server handles a single request, so they are all considered the busiest.

Constraints:

1 <= k <= 10^5
1 <= arrival.length, load.length <= 10^5
arrival.length == load.length
1 <= arrival[i], load[i] <= 10^9
arrival is strictly increasing.

Solution

class Solution {
    static class Server {
        final int id;
        final int busyTime;

        public Server(int id, int busyTime) {
            this.id = id;
            this.busyTime = busyTime;
        }
    }

    public List<Integer> busiestServers(int k, int[] arrival, int[] load) {
        TreeSet<Integer> available = new TreeSet<>();
        PriorityQueue<Server> busy = new PriorityQueue<>(Comparator.comparingInt(a -> a.busyTime));
        int[] requestCount = new int[k];
        int n = arrival.length;
        for (int id = 0; id < k; id++) {
            available.add(id);
        }
        for (int i = 0; i < n; i++) {
            int defaultServer = (i % k);
            while (!busy.isEmpty() && busy.peek().busyTime <= arrival[i]) {
                Server top = busy.poll();
                available.add(top.id);
            }
            if (available.isEmpty()) {
                continue;
            }
            Integer nextServer = available.ceiling(defaultServer);
            nextServer = (nextServer != null) ? nextServer : available.ceiling(0);
            int requestEnd = arrival[i] + load[i];
            available.remove(nextServer);
            busy.add(new Server(nextServer, requestEnd));
            requestCount[nextServer]++;
        }
        int maxRequests = Integer.MIN_VALUE;
        List<Integer> busiestServers = new ArrayList<>();
        for (int id = 0; id < k; id++) {
            maxRequests = Math.max(maxRequests, requestCount[id]);
        }
        for (int id = 0; id < k; id++) {
            if (requestCount[id] == maxRequests) {
                busiestServers.add(id);
            }
        }
        return busiestServers;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).