<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 n;
int mem[500003];
int getindex(vector<vector<int>>&arr,int target,int l){
    int r= n-1;
    int res=n+8;
    while(l<=r){
        int mid=l+(r-l)/2;
        if(arr[mid][0]>=target){
            res=mid;
            r=mid-1;

        }
        else{
            l=mid+1;
        }
    }
    return res;

                                                                              //binary search
}
int solve(vector<vector<int>>&arr,int id){
    
    if(id>=n){
        return 0;
        
    }
    if(mem[id]!=-1){
        return mem[id];
    }
    int next=getindex(arr,arr[id][1],id+1);///imp question
    int taken=arr[id][2]+solve(arr,next);
    int nottaken=solve(arr,id+1);
    return mem[id]=max(taken,nottaken);


}
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        this->n=startTime.size();
        memset(mem,-1,sizeof(mem));
        vector<vector<int>>arr(n,vector<int>(3,0));
        for(int i=0;i<n;i++){
        arr[i][0]=startTime[i];
        arr[i][1]=endTime[i];
        arr[i][2]=profit[i];

    }
    sort(arr.begin(),arr.end());
    return solve(arr,0);
        
    }
};
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