<p>We have <code>n</code> jobs, where every job is scheduled to be done from <code>startTime[i]</code> to <code>endTime[i]</code>, obtaining a profit of <code>profit[i]</code>.</p>

<p>You&#39;re given the <code>startTime</code>, <code>endTime</code> and <code>profit</code> arrays, return the maximum profit you can take such that there are no two jobs in the subset with overlapping time range.</p>

<p>If you choose a job that ends at time <code>X</code> you will be able to start another job that starts at time <code>X</code>.</p>

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

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/10/10/sample1_1584.png" style="width: 380px; height: 154px;" /></strong></p>

<pre>
<strong>Input:</strong> startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]
<strong>Output:</strong> 120
<strong>Explanation:</strong> The subset chosen is the first and fourth job. 
Time range [1-3]+[3-6] , we get profit of 120 = 50 + 70.
</pre>

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

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/10/10/sample22_1584.png" style="width: 600px; height: 112px;" /> </strong></p>

<pre>
<strong>Input:</strong> startTime = [1,2,3,4,6], endTime = [3,5,10,6,9], profit = [20,20,100,70,60]
<strong>Output:</strong> 150
<strong>Explanation:</strong> The subset chosen is the first, fourth and fifth job. 
Profit obtained 150 = 20 + 70 + 60.
</pre>

<p><strong class="example">Example 3:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/10/10/sample3_1584.png" style="width: 400px; height: 112px;" /></strong></p>

<pre>
<strong>Input:</strong> startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4]
<strong>Output:</strong> 6
</pre>

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

<ul>
	<li><code>1 &lt;= startTime.length == endTime.length == profit.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= startTime[i] &lt; endTime[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= profit[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        vector<vector<int>> jobs;
        // storing job's details into one list 
        // this will help in sorting the jobs while maintaining the other parameters
        for (int i = 0; i < profit.size(); i++) {
            jobs.push_back({startTime[i], endTime[i], profit[i]});
        }
        sort(jobs.begin(), jobs.end());
        int n = jobs.size(), maxProfit = 0;
        // min heap having {endTime, profit}
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;
        
        for (int i = 0; i < n; i++) {
            int start = jobs[i][0], end = jobs[i][1], profit = jobs[i][2];
            
            // keep popping while the heap is not empty and jobs are not conflicting
            // update the value of maxProfit
            // eg: m = 50, q  4, 60
            //                5, 90
            //                6, 120
            while (!pq.empty() && start >= pq.top()[0]) { // at this time, we make sure that the pq.top()
                maxProfit = max(maxProfit, pq.top()[1]);  // profit is valid for current index i.
                pq.pop();
            }
            
            // push the job with combined profit
            // if no non-conflicting job is present maxProfit will be 0
            pq.push({end, profit + maxProfit});
        }
        
        // update the value of maxProfit by comparing with
        // profit of jobs that exits in the heap
        while (!pq.empty()) {
            maxProfit = max(maxProfit, pq.top()[1]);
            pq.pop();
        }
        return maxProfit;
    }
};
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