<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://assets.leetcode.com/uploads/2020/09/08/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>

class Solution:
    def busiestServers(self, k: int, arrival: List[int], load: List[int]) -> List[int]:
        available_server=list(range(k))
        busy_server=[]
        count=[0 for _ in range(k)] 
        for i,start in enumerate(arrival):
            while busy_server and busy_server[0][0]<=start:
                endTime,serverId=heapq.heappop(busy_server)
                heapq.heappush(available_server,i+(serverId-i)%k)       
            if available_server:
                serverId=heapq.heappop(available_server)%k
                heapq.heappush(busy_server,(start+load[i],serverId))
                count[serverId]+=1   
        max_job=max(count)
        return [i for i,n in enumerate(count) if n==max_job]
37500