How do you solve Process Tasks Using Servers on LeetCode?

Process Tasks Using Servers (LeetCode #1882) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m log n) time complexity and O(n) space complexity. Key insight: Using priority queues allows for efficient management of server availability.

What is the brute force solution for Process Tasks Using Servers?

The brute force approach for Process Tasks Using Servers is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each server for availability for every task. This is straightforward but inefficient, as it requires scanning through the servers repeatedly. The optimal Optimal Solution approach improves this to O(m log n).

What algorithm or data structure is used to solve Process Tasks Using Servers?

Process Tasks Using Servers uses the following concepts: Array, Heap (Priority Queue). The recommended patterns to study are: Priority Queue, Greedy Algorithms.

What are the interview tips for Process Tasks Using Servers?

Explain your thought process clearly, especially how you manage server states. Discuss edge cases, such as when all servers are busy. Practice coding priority queues, as they are common in scheduling problems.

What are common mistakes when solving Process Tasks Using Servers?

Not updating the time correctly when no servers are free. Forgetting to manage the busy servers queue properly.

#1882

Process Tasks Using Servers

Medium

ArrayHeap (Priority Queue)Priority QueueGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m log n)
Space	O(1)	O(n)

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Intuition

Time O(m log n)Space O(n)

The optimal solution uses two priority queues to efficiently manage free and busy servers. This allows us to quickly assign tasks to the most suitable server without scanning all servers repeatedly.

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Algorithm

3 steps

1Step 1: Initialize two priority queues: one for free servers (sorted by weight and index) and one for busy servers (sorted by finish time).
2Step 2: For each task, check if any servers have finished processing and update the free servers queue.
3Step 3: If a free server is available, assign the task to it and update its busy status. If no servers are free, wait until the next server becomes available.

solution.py24 lines

1import heapq
2
3def assignTasks(servers, tasks):
4    n, m = len(servers), len(tasks)
5    ans = [-1] * m
6    free_servers = [(weight, i) for i, weight in enumerate(servers)]
7    heapq.heapify(free_servers)
8    busy_servers = []
9    time = 0
10
11    for j in range(m):
12        while busy_servers and busy_servers[0][0] <= time:
13            finish_time, server_index = heapq.heappop(busy_servers)
14            heapq.heappush(free_servers, (servers[server_index], server_index))
15
16        if len(free_servers) > 0:
17            weight, server_index = heapq.heappop(free_servers)
18            ans[j] = server_index
19            heapq.heappush(busy_servers, (time + tasks[j], server_index))
20        else:
21            time = busy_servers[0][0]
22            j -= 1  # Re-check the same task
23
24    return ans

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Complexity note: This complexity arises from the use of priority queues, which allow for efficient insertion and extraction of servers, leading to logarithmic time complexity for each task.

1Using priority queues allows for efficient management of server availability.
2Tasks must be assigned in the order they arrive, which requires careful tracking of time.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.