<p>You are given two <strong>0-indexed</strong> integer arrays <code>servers</code> and <code>tasks</code> of lengths <code>n</code>​​​​​​ and <code>m</code>​​​​​​ respectively. <code>servers[i]</code> is the <strong>weight</strong> of the <code>i<sup>​​​​​​th</sup></code>​​​​ server, and <code>tasks[j]</code> is the <strong>time needed</strong> to process the <code>j<sup>​​​​​​th</sup></code>​​​​ task <strong>in seconds</strong>.</p>

<p>Tasks are assigned to the servers using a <strong>task queue</strong>. Initially, all servers are free, and the queue is <strong>empty</strong>.</p>

<p>At second <code>j</code>, the <code>j<sup>th</sup></code> task is <strong>inserted</strong> into the queue (starting with the <code>0<sup>th</sup></code> task being inserted at second <code>0</code>). As long as there are free servers and the queue is not empty, the task in the front of the queue will be assigned to a free server with the <strong>smallest weight</strong>, and in case of a tie, it is assigned to a free server with the <strong>smallest index</strong>.</p>

<p>If there are no free servers and the queue is not empty, we wait until a server becomes free and immediately assign the next task. If multiple servers become free at the same time, then multiple tasks from the queue will be assigned <strong>in order of insertion</strong> following the weight and index priorities above.</p>

<p>A server that is assigned task <code>j</code> at second <code>t</code> will be free again at second <code>t + tasks[j]</code>.</p>

<p>Build an array <code>ans</code>​​​​ of length <code>m</code>, where <code>ans[j]</code> is the <strong>index</strong> of the server the <code>j<sup>​​​​​​th</sup></code> task will be assigned to.</p>

<p>Return <em>the array </em><code>ans</code>​​​​.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> servers = [3,3,2], tasks = [1,2,3,2,1,2]
<strong>Output:</strong> [2,2,0,2,1,2]
<strong>Explanation: </strong>Events in chronological order go as follows:
- At second 0, task 0 is added and processed using server 2 until second 1.
- At second 1, server 2 becomes free. Task 1 is added and processed using server 2 until second 3.
- At second 2, task 2 is added and processed using server 0 until second 5.
- At second 3, server 2 becomes free. Task 3 is added and processed using server 2 until second 5.
- At second 4, task 4 is added and processed using server 1 until second 5.
- At second 5, all servers become free. Task 5 is added and processed using server 2 until second 7.</pre>

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

<pre>
<strong>Input:</strong> servers = [5,1,4,3,2], tasks = [2,1,2,4,5,2,1]
<strong>Output:</strong> [1,4,1,4,1,3,2]
<strong>Explanation: </strong>Events in chronological order go as follows: 
- At second 0, task 0 is added and processed using server 1 until second 2.
- At second 1, task 1 is added and processed using server 4 until second 2.
- At second 2, servers 1 and 4 become free. Task 2 is added and processed using server 1 until second 4. 
- At second 3, task 3 is added and processed using server 4 until second 7.
- At second 4, server 1 becomes free. Task 4 is added and processed using server 1 until second 9. 
- At second 5, task 5 is added and processed using server 3 until second 7.
- At second 6, task 6 is added and processed using server 2 until second 7.
</pre>

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

<ul>
	<li><code>servers.length == n</code></li>
	<li><code>tasks.length == m</code></li>
	<li><code>1 &lt;= n, m &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>1 &lt;= servers[i], tasks[j] &lt;= 2 * 10<sup>5</sup></code></li>
</ul>

class Solution:
    def assignTasks(self, servers: List[int], tasks: List[int]) -> List[int]:
        server_heap=[(weight,id) for id,weight in enumerate(servers)]
        heapify(server_heap)
        time = 0
        res=[]
        processing=[]
        for task in tasks:
            
            while processing and processing[0][0]<=time:
                _,weight,id=heapq.heappop(processing)
                heapq.heappush(server_heap,(weight,id))


            if len(server_heap)>0:
                weight,id= heapq.heappop(server_heap)
                heapq.heappush(processing,(time+task,weight,id))
                res.append(id)
            else:
                finish_time,weight,id = heapq.heappop(processing)
                heapq.heappush(processing,(finish_time+task,weight,id))
                res.append(id)

            time+=1
        return res
            





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