<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:

    static bool mycompare(const vector<int>&v1, const vector<int>&v2)
    {
        return v1[0]<v2[0];
    }

    int BS(vector<vector<int>>&vec, int el, int l, int r)
    {
        int ans=r+1;
        while(l<=r)
        {
            int mid=(l+r)/2;
            if(vec[mid][0]>=el)
            {
                ans=min(ans,mid);
                r=mid-1;
            }
            else l=mid+1;
            
        }
        return ans;
    }

    int MP(int i, vector<vector<int>>&temp,int n, vector<int>&dp)
    {
        if(i==n) return 0;
        if(dp[i]!=-1) return dp[i];
        int Pick=0;
        int noPick=0;
        noPick=MP(i+1,temp,n,dp);
        int idx=BS(temp,temp[i][1],i+1,n-1);
        Pick=temp[i][2]+MP(idx,temp,n,dp);
        return dp[i]=max(noPick,Pick);
    }

    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        vector<vector<int>>temp;
        int n=profit.size();
        for(int i=0;i<n;i++)
        {
            temp.push_back({startTime[i],endTime[i],profit[i]});
        }
        vector<int>dp(n,-1);
        sort(temp.begin(),temp.end(),mycompare);
        return MP(0,temp,n,dp);
    }
};
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