Find Servers That Handled Most Number of Requests

This page explains Java solution to problem Find Servers That Handled Most Number of Requests using TreeMap data structure.

Problem Statement

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]


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import java.util.*;

class FindServersThatHandledMostNumberOfRequests {
    public List<Integer> busiestServers(int k, int[] arrival, int[] load) {
        TreeSet<Integer> available = new TreeSet<>();
        for(int server = 0; server < k; server++) available.add(server);

        //[server, busyUntil order by Asc]
        PriorityQueue<int[]> pq = new PriorityQueue<>(new Comparator<int[]>(){
            public int compare(int[] x, int[] y) {
                int cmp =[1], y[1]);
                if(cmp == 0) return[0], y[0]);
                return cmp;

        int[] count = new int[k];
        int max = 0;
        for(int i = 0; i < arrival.length; i++) {
            int serverIndex = i % k;
            int arrivalTime = arrival[i];
            while(!pq.isEmpty() && pq.peek()[1] <= arrivalTime) {
            Integer availableServer = available.ceiling(serverIndex);

            //wrapping around
            if(availableServer == null) availableServer = available.ceiling(0);

            if(availableServer != null) {
                int busyUntil = arrivalTime + load[i];
                pq.offer(new int[]{availableServer, busyUntil});
                max = Math.max(max, count[availableServer]);

        List<Integer> res = new LinkedList<>();
        for(int i = 0; i < k; i++) if(count[i] == max) res.add(i);

        return res;

Time Complexity

O(N * log K)) Where
N is number of request in an input array arrival
K is number of servers to handle request

Space Complexity

O(K) Where
K is number of servers to handle request