<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:
    vector<int> dp;
    Solution(): dp(50001,-1){}
    int  find_NEXT(int i,int n,int target,vector<vector<int>> &v)
    {
        int result=-1;
        while(i<=n)
        {
            int mid=(i+n)/2;
            if(v[mid][1]<=target)
            {
                result=mid;
                i=mid+1;
            }
            else
            {
                n=mid-1;
            }
        }
        return result;
    }
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        int n=startTime.size();
       vector<vector<int>> v;
       for(int i=0;i<startTime.size();i++)
       {
            v.push_back({startTime[i],endTime[i],profit[i]});
       } 
       
       auto comp=[&](vector<int> &v1,vector<int> &v2)
       {
        return v1[1]<v2[1];
       };
       sort(v.begin(),v.end(),comp);
       
        vector<int> dp(n);

        dp[0]=v[0][2];
        for(int i=1;i<n;i++)
        {
            int prev=0;
            int lastjobindex=find_NEXT(0,i-1,v[i][0],v);
            if(lastjobindex!=-1)
            {
                prev=dp[lastjobindex];
            }
            dp[i]=max(dp[i-1],prev+v[i][2]);
        }
        return dp[n-1];
    }
};
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