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Parallel Task Completion Order

Run tasks concurrently with at most k at a time and report the order in which they finish.

Чему вы научитесь

  • Run tasks with a limit on parallelism
  • Simulate time with a priority queue
  • Break ties between tasks that finish together

Условие

Показать перевод

With async/await you can start several tasks and let them run at the same time. A SemaphoreSlim often caps how many run at once.

Tasks numbered 0 to n − 1 take durations[i] time units. At time 0 start as many tasks as the limit allows, in index order. Whenever tasks finish, record them (if several finish at the same moment, lower index first) and immediately start the next waiting tasks, in index order, to fill the free slots.

Implement parallel_task_order(durations, limit).

Input

durations: How long each task takes.

limit: The maximum number of tasks running at the same time.

Output

Return the task indices in the order they finish.

Ограничения

1 ≤ n ≤ 100,000

1 ≤ durations[i] ≤ 10^6; 1 ≤ limit ≤ n

Примеры

Пример 1
Ввод:
limit = 2
durations = [4, 2, 3, 1]

Ожидаемый вывод:
expected = [1, 0, 2, 3]
Ввод:
long long limit = 2;
std::vector<long long> durations = {4, 2, 3, 1};

Ожидаемый вывод:
std::vector<long long> expected = {1, 0, 2, 3};
Ввод:
long limit = 2;
long[] durations = new long[] {4, 2, 3, 1};

Ожидаемый вывод:
long[] expected = new long[] {1, 0, 2, 3};

Пояснение

Tasks 0 and 1 start. Task 1 ends at 2 and task 2 starts (ends at 5). Task 0 ends at 4 and task 3 starts (ends at 5). Tasks 2 and 3 finish together, lower index first: [1, 0, 2, 3].

Пример 2
Ввод:
limit = 3
durations = [3, 1, 2]

Ожидаемый вывод:
expected = [1, 2, 0]
Ввод:
long long limit = 3;
std::vector<long long> durations = {3, 1, 2};

Ожидаемый вывод:
std::vector<long long> expected = {1, 2, 0};
Ввод:
long limit = 3;
long[] durations = new long[] {3, 1, 2};

Ожидаемый вывод:
long[] expected = new long[] {1, 2, 0};

Пояснение

All tasks run at once and finish at 1, 2 and 3: [1, 2, 0].

Как подступиться

Сначала попробуйте сами, затем открывайте подсказки по одной.

Подсказка 1

Keep the running tasks ordered by finish time, then by index.

Подсказка 2

Each step jumps to the next finish time and collects every task ending then.

Подсказка 3

Refill free slots with the next waiting tasks in index order.

Ожидаемая сложность

O(n log limit) time, O(limit) extra space

Ссылки

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