<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:
    vector<int> getOrder(vector<vector<int>>& tasks) 
    {
        vector<pair<vector<int>, int >> tasksI;

        for (unsigned i = 0; i < tasks.size(); ++i)
        {
            tasksI.push_back({ tasks.at(i), i });
        }

        std::sort(tasksI.begin(), tasksI.end());

        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>>pqueue;

        std::vector<int> processed;

        long long int cycle = tasksI.begin()->first.at(0);
        std::size_t i = 0;
        int currentTaskIndex = -1;

        auto tasksNo = tasksI.size();

        while (true)
        {
            // Check if there are elements to insert in queue
            while (i < tasksI.size() && cycle >= tasksI[i].first.at(0))
            {
                pqueue.push(std::make_pair(tasksI[i].first.at(1), tasksI[i].second));
                i++;
            }
           
            if (!pqueue.empty())
            {
                auto it = pqueue.top();
                    
                cycle += it.first;
                processed.emplace_back(it.second);

                if (processed.size() == tasksNo)
                {
                    break;
                }

                pqueue.pop();
            }
                            else
            {
                // Queue is empty
                cycle = tasksI[i].first[0];
            }
            
        }

        return processed;
}
};
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