<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>

from sortedcontainers import SortedList


class Solution:
    def busiestServers(self, k: int, arrival: List[int], load: List[int]) -> List[int]:
        # Input: m = 3, queue_time = [1,2,3,7], processing_time = [5,4,3,2], k = 2
        # 0 1 2 3 4 5 6 7 8
        #     -----------
        #       ---------
        #         -------
        #                 --
        # 0: Task 0 Task 3
        # 1: Task 1
        # 2: Task 2

        # Naive solution:
        # O(N*k)
        # For each i, queue_at, process_time in queue:
        # 1. check if (i % k) -> (i-1 %k) server is available.
        # 2. If none available, drop it
        # 3. update counter for used server, update total requests processed, update server -> busy_until


        # O(N*logK) 
        # busy -> min heap of (busy_until, indexer ID)
        # free -> sorted list of (indexer ID), do wrap-around binary search to find min server >= (i % k)
        # binary search (if element exists return, else choose 0-th)
        
        processed_counter = defaultdict(int)
        busy = []
        free = SortedList([i for i in range(k)])
        for task_id, queue_at in enumerate(arrival):
            process_time = load[task_id]
            # free up server that no longer busy
            while busy:
                busy_until, busy_server_index = busy[0]
                if busy_until > queue_at:
                    break
                heappop(busy)
                free.add(busy_server_index)
            if not free:
                continue
            # choose a free server
            next_index = free.bisect_right(task_id%k - 1)
            if next_index < len(free):
                to_free_server_index = free[next_index]
            else:
                to_free_server_index = free[0]
            processed_counter[to_free_server_index] += 1
            free.remove(to_free_server_index)
            heappush(busy, (queue_at+process_time, to_free_server_index))
        most_processed = max(processed_counter.values())
        # print(most_processed, dict(processed_counter))
        return [i for i, n in processed_counter.items() if n == most_processed]






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