Process Tasks Using Servers - Leetcode Solution

Problem Summary:

The problem "Process Tasks Using Servers" on LeetCode is a task scheduling problem that involves assigning tasks to available servers in order to minimize the time taken to process all the tasks. The problem statement is as follows:

Given two arrays - servers and tasks, with servers[i] representing the weight of the i-th server and tasks[j] representing the time needed to process the j-th task, you have to assign the tasks to the servers in such a way that the total time taken to process all the tasks is minimized.

Constraints:

1. 1 ≤ servers.length ≤ 2 * 10^5
2. 1 ≤ tasks.length ≤ 2 * 10^5
3. 1 ≤ servers[i], tasks[j] ≤ 2 * 10^5

Solution:

The problem can be solved using a priority queue and the two pointer technique. We can first create a priority queue of available servers sorted by their weights in ascending order. Then, we can iterate through the tasks and for each task, we can pop the server with the lowest weight from the priority queue and assign it to the task. We can then update the total time taken to process the tasks as the sum of the time taken to process the task and the weight of the server. Once we have assigned all the tasks, we return the total time taken.

Algorithm:

1. Define a priority queue of available servers sorted by their weights in ascending order.
2. Define a two-pointer 'i' and 'j', starting at 0.
3. Define a list 'result' to store the assigned server for each task.
4. Define a variable 'time_elapsed' to store the time taken to process all the tasks.
5. Iterate through the tasks from 0 to n-1: a. While the priority queue is not empty and i < n: i. Pop the server with the lowest weight from the priority queue. ii. Assign the server to the current task. iii. Append the assigned server to the result list. iv. Update the time elapsed as the sum of the time taken to process the task and the weight of the server. v. Increment i. b. If the priority queue is empty, increment j until we find a server that will be available to process the next task.
6. Return the list result.

Pseudo Code:

servers_queue = [] // priority queue of servers sorted by their weights n = length(tasks) result = [] time_elapsed = 0 i = j = 0

for server in servers: add server to servers_queue

for task in tasks: while servers_queue is not empty and i < n: assigned_server = pop the server with lowest weight from servers_queue result[task] = assigned_server time_elapsed += assigned_server + tasks[task] i += 1

if servers_queue is empty: while j < n and j <= i: if servers[j] is available: add servers[j] to servers_queue j += 1

return result

Complexity Analysis:

Time Complexity: O(n*log(n)) where n is the total number of tasks. The priority queue is updated n times and each update takes log(n) time, so the time complexity of this algorithm is log(n) for each of the n tasks.

Space Complexity: O(n) where n is the total number of tasks. We need to store the assigned server for each task, which would take n space. Additionally, we need space to store the priority queue of servers, which would take O(n) space as well.

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