<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 jobData
{
    public:
    int start;
    int end;
    int reward;
    jobData(int s,int e,int r)
    {
        start=s;
        end=e;
        reward=r;
    }
};

int BSCeil(vector<jobData>&vec,int lo,int hi,int time)
{
    int ansIdx=vec.size();
    
    while(lo<=hi)
    {
        int mid=lo+(hi-lo)/2;
        if(vec[mid].start>=time)
        {
            ansIdx=mid;
            hi=mid-1;
        }
        else
        {
            lo=mid+1;
        }
    }
    return ansIdx;
}

int helper(vector<jobData>&vec,int idx,int prevEnd,vector<int>&dp)
{

    //this func would provide the profit one can get by performing jobs
    //with a option of performing or not for the "idx" job
    if(idx==vec.size())
    {
        return 0;
    }
    if(dp[idx]!=-1)
    {
        return dp[idx];
    }
    
    if(prevEnd<=vec[idx].start)
    {
        //we can take current job
        int k=BSCeil(vec,idx+1,vec.size()-1,vec[idx].end);
        int taken=vec[idx].reward+helper(vec,k,vec[idx].end,dp);
        int notTaken=helper(vec,idx+1,prevEnd,dp);
        return dp[idx]=max(taken,notTaken);
    }
    else
    {
        return dp[idx]=helper(vec,idx+1,prevEnd,dp);
    }

}

class Solution {
public:
    static bool compare(jobData j1,jobData j2)
    {
        return j1.start<j2.start;
    }
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        int n=startTime.size();
        if(n==1)return profit[0];
        vector<jobData>vec;
        for(int i=0;i<n;i++)
        {

            vec.push_back(jobData(startTime[i],endTime[i],profit[i]));
        }
        vector<int>dp(n+1,-1);
        sort(vec.begin(),vec.end(),compare);
        int finalAns=helper(vec,0,0,dp);
        return finalAns;
        
    }
};
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