<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 {
private:
int n;
int memo[50001];
int next_job(vector<int>& startTime,int end)
{
    int l=0;
    int r=n-1;
    while(l<=r)
    {
       int mid=l+(r-l)/2;    
       if(end>startTime[mid])
       l=mid+1;
       else
       r=mid-1; 
    }
    return l;   
}
void sort(vector<int>& startTime,vector<int>& endTime, vector<int>& profit )
    {
        int pr,j,s,e;
        for(int i=1;i<profit.size();i++)
        {
            pr=profit[i];
            s=startTime[i];
            e=endTime[i];
            j=i-1;
            while(j>=0&&startTime[j]>s)
            {
                profit[j+1]=profit[j];
                startTime[j+1]=startTime[j];
                endTime[j+1]=endTime[j];
                j--;
            }
                profit[j+1]=pr;
                startTime[j+1]=s;
                endTime[j+1]=e;
        }
    }
  int solve(vector<int>& startTime,vector<int>& endTime, vector<int>& profit,int i ) 
  {
    if(i>=n)
    return 0;
    if(memo[i]!=-1)
    return memo[i];
    int next=next_job(startTime,endTime[i]);
    int taken=profit[i]+solve(startTime,endTime,profit,next );
    int not_taken=solve(startTime,endTime,profit,i+1);
    memo[i]=max(taken,not_taken);
    return memo[i];
  } 
public:
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        n=profit.size();
        memset(memo,-1,sizeof(memo));
        sort(startTime,endTime,profit );
        return solve(startTime,endTime, profit,0) ;
    }
};
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