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)
func busiestServers(k int, arrival, load []int) (ans []int) {
	free := redblacktree.NewWithIntComparator()
	for i := 0; i < k; i++ {
		free.Put(i, nil)
	}
	busy := hp{}
	cnt := make([]int, k)
	for i, t := range arrival {
		for len(busy) > 0 && busy[0].end <= t {
			free.Put(busy[0].server, nil)
			heap.Pop(&busy)
		}
		if free.Size() == 0 {
			continue
		}
		p, _ := free.Ceiling(i % k)
		if p == nil {
			p = free.Left()
		}
		server := p.Key.(int)
		cnt[server]++
		heap.Push(&busy, pair{t + load[i], server})
		free.Remove(server)
	}
	mx := slices.Max(cnt)
	for i, v := range cnt {
		if v == mx {
			ans = append(ans, i)
		}
	}
	return
}

type pair struct{ end, server int }
type hp []pair

func (h hp) Len() int           { return len(h) }
func (h hp) Less(i, j int) bool { return h[i].end < h[j].end }
func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(v any)        { *h = append(*h, v.(pair)) }
func (h *hp) Pop() any          { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }