You are given `n`

tasks labeled from `0`

to `n - 1`

represented by a 2D integer array `tasks`

, where `tasks[i] = [enqueueTime`_{i}, processingTime_{i}]

means that the `i`^{th}

task will be available to process at `enqueueTime`_{i}

and will take `processingTime`_{i}

to finish processing.

You have a single-threaded CPU that can process **at most one** task at a time and will act in the following way:

- If the CPU is idle and there are no available tasks to process, the CPU remains idle.
- If the CPU is idle and there are available tasks, the CPU will choose the one with the
**shortest processing time**. If multiple tasks have the same shortest processing time, it will choose the task with the smallest index. - Once a task is started, the CPU will
**process the entire task** without stopping. - The CPU can finish a task then start a new one instantly.

Return *the order in which the CPU will process the tasks.*

**Example 1:**

**Input:** tasks = [[1,2],[2,4],[3,2],[4,1]]
**Output:** [0,2,3,1]
**Explanation: **The events go as follows:
- At time = 1, task 0 is available to process. Available tasks = {0}.
- Also at time = 1, the idle CPU starts processing task 0. Available tasks = {}.
- At time = 2, task 1 is available to process. Available tasks = {1}.
- At time = 3, task 2 is available to process. Available tasks = {1, 2}.
- Also at time = 3, the CPU finishes task 0 and starts processing task 2 as it is the shortest. Available tasks = {1}.
- At time = 4, task 3 is available to process. Available tasks = {1, 3}.
- At time = 5, the CPU finishes task 2 and starts processing task 3 as it is the shortest. Available tasks = {1}.
- At time = 6, the CPU finishes task 3 and starts processing task 1. Available tasks = {}.
- At time = 10, the CPU finishes task 1 and becomes idle.

**Example 2:**

**Input:** tasks = [[7,10],[7,12],[7,5],[7,4],[7,2]]
**Output:** [4,3,2,0,1]
**Explanation****: **The events go as follows:
- At time = 7, all the tasks become available. Available tasks = {0,1,2,3,4}.
- Also at time = 7, the idle CPU starts processing task 4. Available tasks = {0,1,2,3}.
- At time = 9, the CPU finishes task 4 and starts processing task 3. Available tasks = {0,1,2}.
- At time = 13, the CPU finishes task 3 and starts processing task 2. Available tasks = {0,1}.
- At time = 18, the CPU finishes task 2 and starts processing task 0. Available tasks = {1}.
- At time = 28, the CPU finishes task 0 and starts processing task 1. Available tasks = {}.
- At time = 40, the CPU finishes task 1 and becomes idle.

**Constraints:**

`tasks.length == n`

`1 <= n <= 10`^{5}

`1 <= enqueueTime`_{i}, processingTime_{i} <= 10^{9}

**Solution: Simulation w/ Sort + PQ**

Time complexity: O(nlogn)

Space complexity: O(n)

## C++

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// Author: Huahua class Solution { public: vector<int> getOrder(vector<vector<int>>& tasks) { const int n = tasks.size(); for (int i = 0; i < n; ++i) tasks[i].push_back(i); sort(begin(tasks), end(tasks)); // sort by enqueue_time; vector<int> ans; priority_queue<pair<int, int>> q; // {-processing_time, -index} int i = 0; long t = 0; while (ans.size() != n) { // Advance to next enqueue time if q is empty. if (i < n && q.empty() && tasks[i][0] > t) t = tasks[i][0]; // Enqueue all available tasks. while (i < n && tasks[i][0] <= t) { q.emplace(-tasks[i][1], -tasks[i][2]); ++i; } // Extra the top one. t -= q.top().first; ans.push_back(-q.top().second); q.pop(); } return ans; } }; |