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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.

Das lernst du

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

Problem

Übersetzung anzeigen

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.

Einschränkungen

1 ≤ n ≤ 100,000

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

Beispiele

Beispiel 1
Eingabe:
limit = 2
durations = [4, 2, 3, 1]

Erwartete Ausgabe:
expected = [1, 0, 2, 3]
Eingabe:
long long limit = 2;
std::vector<long long> durations = {4, 2, 3, 1};

Erwartete Ausgabe:
std::vector<long long> expected = {1, 0, 2, 3};
Eingabe:
long limit = 2;
long[] durations = new long[] {4, 2, 3, 1};

Erwartete Ausgabe:
long[] expected = new long[] {1, 0, 2, 3};

Erläuterung

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].

Beispiel 2
Eingabe:
limit = 3
durations = [3, 1, 2]

Erwartete Ausgabe:
expected = [1, 2, 0]
Eingabe:
long long limit = 3;
std::vector<long long> durations = {3, 1, 2};

Erwartete Ausgabe:
std::vector<long long> expected = {1, 2, 0};
Eingabe:
long limit = 3;
long[] durations = new long[] {3, 1, 2};

Erwartete Ausgabe:
long[] expected = new long[] {1, 2, 0};

Erläuterung

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

So gehst du vor

Versuch es zuerst allein und öffne dann einen Hinweis nach dem anderen.

Hinweis 1

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

Hinweis 2

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

Hinweis 3

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

Erwartete Komplexität

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

Referenzen

Diese Aufgabe wurde für CodeUtility angepasst. Beachte die oben genannten Vorgaben und Einschränkungen.

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