<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>& start, vector<int>& end, vector<int>& p) {
        //ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
        int l=start.size();vector<vector<int>> all; //for sorting three vectors
        for(int i=0;i<l;i++) all.push_back({start[i],end[i],p[i]});
        sort(all.begin(),all.end());
        for(int i=0;i<l;i++){
            start[i]=all[i][0];end[i]=all[i][1];p[i]=all[i][2];
        }
        vector<int> maxP(l,0);//maxP[i] is the max profit if the input list only include jobs from i to l-1
        for(int i=l-1;i>=0;i--){ //iterate the job list backwardly to build up the answer vector 
            //maxP[i] equals to maxP[i+1] if job i is not taken
            //otherwise maxP[i] is the p[i] + maxP[j] j is the first (earliest) job that has no time conflict with i 
            //so we need to compare this two situations to achieve maxprofit
            if(i==l-1) {maxP[i]=p[i];continue;} //i is the only job on the list
            int j=i+1; //find j by checking following jobs that cause time conflicts
            while(start[j]<end[i]) { 
                if(end[j]<=end[i]) if(p[j]>p[i]) j=l-1;// if a more profitable job contained in i's time window, i must be abandonded
                j++; if(j==l) break; 
            } 
            if(j==i+1) {maxP[i]=p[i]+maxP[i+1];continue;} // i has no confilcts with all the following jobs
            int maxA=maxP[i+1]; //max profit if i is not taken
            int maxB;// max profit if i is taken
            if(j==l) maxB= p[i];// i has conflicts with all following jobs, so it is the only job to do
            else maxB=p[i] +maxP[j];// take i and continue with jobs with no time conflicts with i
            maxP[i]=max(maxA, maxB);
        }
        return maxP[0];
    }    
};
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