Find Servers That Handled Most Number of Requests
<p>You have <code>k</code> servers numbered from <code>0</code> to <code>k-1</code> that are being used to handle multiple requests simultaneously. Each server has infinite computational capacity but <strong>cannot handle more than one request at a time</strong>. The requests are assigned to servers according to a specific algorithm:</p>

<ul>
	<li>The <code>i<sup>th</sup></code> (0-indexed) request arrives.</li>
	<li>If all servers are busy, the request is dropped (not handled at all).</li>
	<li>If the <code>(i % k)<sup>th</sup></code> server is available, assign the request to that server.</li>
	<li>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 <code>i<sup>th</sup></code> server is busy, try to assign the request to the <code>(i+1)<sup>th</sup></code> server, then the <code>(i+2)<sup>th</sup></code> server, and so on.</li>
</ul>

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

<p>Return <em>a list containing the IDs (0-indexed) of the <strong>busiest server(s)</strong></em>. You may return the IDs in any order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1600-1699/1606.Find%20Servers%20That%20Handled%20Most%20Number%20of%20Requests/images/load-1.png" style="width: 389px; height: 221px;" />
<pre>
<strong>Input:</strong> k = 3, arrival = [1,2,3,4,5], load = [5,2,3,3,3] 
<strong>Output:</strong> [1] 
<strong>Explanation:</strong> 
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&#39;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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> k = 3, arrival = [1,2,3,4], load = [1,2,1,2]
<strong>Output:</strong> [0]
<strong>Explanation:</strong> 
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.
</pre>

<p><strong class="example">Example 3:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= arrival.length, load.length &lt;= 10<sup>5</sup></code></li>
	<li><code>arrival.length == load.length</code></li>
	<li><code>1 &lt;= arrival[i], load[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>arrival</code> is <strong>strictly increasing</strong>.</li>
</ul>
Greedy; Array; Ordered Set; Heap (Priority Queue)
class Solution {
    public List<Integer> busiestServers(int k, int[] arrival, int[] load) {
        int[] cnt = new int[k];
        PriorityQueue<int[]> busy = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
        TreeSet<Integer> free = new TreeSet<>();
        for (int i = 0; i < k; ++i) {
            free.add(i);
        }
        for (int i = 0; i < arrival.length; ++i) {
            int start = arrival[i];
            int end = start + load[i];
            while (!busy.isEmpty() && busy.peek()[0] <= start) {
                free.add(busy.poll()[1]);
            }
            if (free.isEmpty()) {
                continue;
            }
            Integer server = free.ceiling(i % k);
            if (server == null) {
                server = free.first();
            }
            ++cnt[server];
            busy.offer(new int[] {end, server});
            free.remove(server);
        }
        int mx = 0;
        for (int v : cnt) {
            mx = Math.max(mx, v);
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < k; ++i) {
            if (cnt[i] == mx) {
                ans.add(i);
            }
        }
        return ans;
    }
}