問題
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.
制約
1 ≤ n ≤ 100,000
1 ≤ durations[i] ≤ 10^6; 1 ≤ limit ≤ n
例
入力：
limit = 2
durations = [4, 2, 3, 1]
期待される出力：
expected = [1, 0, 2, 3]入力：
long long limit = 2;
std::vector<long long> durations = {4, 2, 3, 1};
期待される出力：
std::vector<long long> expected = {1, 0, 2, 3};入力：
long limit = 2;
long[] durations = new long[] {4, 2, 3, 1};
期待される出力：
long[] expected = new long[] {1, 0, 2, 3};解説
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].
入力：
limit = 3
durations = [3, 1, 2]
期待される出力：
expected = [1, 2, 0]入力：
long long limit = 3;
std::vector<long long> durations = {3, 1, 2};
期待される出力：
std::vector<long long> expected = {1, 2, 0};入力：
long limit = 3;
long[] durations = new long[] {3, 1, 2};
期待される出力：
long[] expected = new long[] {1, 2, 0};解説
All tasks run at once and finish at 1, 2 and 3: [1, 2, 0].
解き方のヒント
まずは自分で考えてから、ヒントを 1 つずつ開いてみましょう。
ヒント 1
Keep the running tasks ordered by finish time, then by index.
ヒント 2
Each step jumps to the next finish time and collects every task ending then.
ヒント 3
Refill free slots with the next waiting tasks in index order.
目安の計算量
O(n log limit) time, O(limit) extra space
参考資料
この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。