Class Solution
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public final class Solution
1882 - Process Tasks Using Servers\.
Medium
You are given two 0-indexed integer arrays
servers
andtasks
of lengthsn
andm
respectively.servers[i]
is the weight of the <code>i<sup>th</sup></code> server, andtasks[j]
is the time needed to process the <code>j<sup>th</sup></code> task in seconds.Tasks are assigned to the servers using a task queue. Initially, all servers are free, and the queue is empty.
At second
j
, the <code>j<sup>th</sup></code> task is inserted into the queue (starting with the <code>0<sup>th</sup></code> task being inserted at second0
). 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 smallest weight , and in case of a tie, it is assigned to a free server with the smallest index.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 in order of insertion following the weight and index priorities above.
A server that is assigned task
j
at secondt
will be free again at secondt + tasks[j]
.Build an array
ans
of lengthm
, whereans[j]
is the index of the server the <code>j<sup>th</sup></code> task will be assigned to.Return the array
ans
.Example 1:
Input: servers = 3,3,2, tasks = 1,2,3,2,1,2
Output: 2,2,0,2,1,2
Explanation: 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.
Example 2:
Input: servers = 5,1,4,3,2, tasks = 2,1,2,4,5,2,1
Output: 1,4,1,4,1,3,2
Explanation: 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.
Constraints:
servers.length == n
tasks.length == m
<code>1 <= n, m <= 2 * 10<sup>5</sup></code>
<code>1 <= serversi, tasksj<= 2 * 10<sup>5</sup></code>
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Constructor Summary
Constructors Constructor Description Solution()
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Method Summary
Modifier and Type Method Description final IntArray
assignTasks(IntArray servers, IntArray tasks)
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Method Detail
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assignTasks
final IntArray assignTasks(IntArray servers, IntArray tasks)
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