Problema
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.
Vincoli
1 ≤ n ≤ 100,000
1 ≤ durations[i] ≤ 10^6; 1 ≤ limit ≤ n
Esempi
Input:
limit = 2
durations = [4, 2, 3, 1]
Output atteso:
expected = [1, 0, 2, 3]Input:
long long limit = 2;
std::vector<long long> durations = {4, 2, 3, 1};
Output atteso:
std::vector<long long> expected = {1, 0, 2, 3};Input:
long limit = 2;
long[] durations = new long[] {4, 2, 3, 1};
Output atteso:
long[] expected = new long[] {1, 0, 2, 3};Spiegazione
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].
Input:
limit = 3
durations = [3, 1, 2]
Output atteso:
expected = [1, 2, 0]Input:
long long limit = 3;
std::vector<long long> durations = {3, 1, 2};
Output atteso:
std::vector<long long> expected = {1, 2, 0};Input:
long limit = 3;
long[] durations = new long[] {3, 1, 2};
Output atteso:
long[] expected = new long[] {1, 2, 0};Spiegazione
All tasks run at once and finish at 1, 2 and 3: [1, 2, 0].
Come affrontarlo
Prova prima da solo, poi apri un suggerimento alla volta.
Suggerimento 1
Keep the running tasks ordered by finish time, then by index.
Suggerimento 2
Each step jumps to the next finish time and collects every task ending then.
Suggerimento 3
Refill free slots with the next waiting tasks in index order.
Complessità attesa
O(n log limit) time, O(limit) extra space
Riferimenti
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