<p>You are given <code>n</code>​​​​​​ tasks labeled from <code>0</code> to <code>n - 1</code> represented by a 2D integer array <code>tasks</code>, where <code>tasks[i] = [enqueueTime<sub>i</sub>, processingTime<sub>i</sub>]</code> means that the <code>i<sup>​​​​​​th</sup></code>​​​​ task will be available to process at <code>enqueueTime<sub>i</sub></code> and will take <code>processingTime<sub>i</sub></code><sub> </sub>to finish processing.</p>

<p>You have a single-threaded CPU that can process <strong>at most one</strong> task at a time and will act in the following way:</p>

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

<p>Return <em>the order in which the CPU will process the tasks.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> tasks = [[1,2],[2,4],[3,2],[4,1]]
<strong>Output:</strong> [0,2,3,1]
<strong>Explanation: </strong>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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> tasks = [[7,10],[7,12],[7,5],[7,4],[7,2]]
<strong>Output:</strong> [4,3,2,0,1]
<strong>Explanation</strong><strong>: </strong>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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>tasks.length == n</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= enqueueTime<sub>i</sub>, processingTime<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
public:
    std::vector<int> getOrder(std::vector<std::vector<int>>& tasks) {
        int n = tasks.size(); //declares an integer n and initializes it with the number of tasks
        std::vector<std::vector<int>> indexedTasks(n);

        for (int i = 0; i < n; ++i) {
            indexedTasks[i] = {tasks[i][0], tasks[i][1], i};
        } // this loop goes through each task, and for each task, it creates a new vector containing the enqueue time, processing time, and the original index of the task, and stores this new vector in indexedTasks

        std::sort(indexedTasks.begin(), indexedTasks.end());
        //sorts indexedTasks in ascending order based on their enqueue time.

        std::priority_queue<std::vector<int>, std::vector<std::vector<int>>, std::greater<>> pq;
        //declares a priority queue pq that stores vectors of integers the priority queue is ordered by std::greater<> which means it’s a min-heap

        std::vector<int> result;
        int index = 0;
        long long current_time = 0; 
        
        while (!pq.empty() || index < n) {
            while (index < n && indexedTasks[index][0] <= current_time) {
                pq.push({indexedTasks[index][1], indexedTasks[index][2]});
                index++;
                //inner loop pushes tasks whose enqueue time is less than or equal to current_time into the priority queue, and increments index
            }
            
            if (pq.empty()) {
                current_time = indexedTasks[index][0];
                continue;
            }
            // priority queue is empty, this code updates current_time to the enqueue time of the next task and skips the rest of the loop.
            
            auto task = pq.top();
            pq.pop();
            
            int processingTime = task[0];
            int originalIndex = task[1];
            
            current_time += processingTime; 
            result.push_back(originalIndex);
            //update current_time by adding the processing time of the task, and add the original index of the task to result
        }
        
        return result;
        //returns result, which is the order of execution of the tasks.
    }
};

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