<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 {
    struct obj{
        int start;
        int end;
        int prof;
    };

    struct comp{
        bool operator()(const obj&a, const obj&b){
            return a.start < b.start;
        }
    };
    struct comp2{
        bool operator()(const obj&a, const obj&b){
            return a.end < b.start;
        }
    };
    

public:


    int solve(int s, vector<obj>&j, vector<int>&dp, vector<int>&startTime){
        
         if(s >= j.size()){
            return 0;
        }
        if(s == j.size()-1){
            return j[s].prof;
        }
     
        if(dp[s] != -1){
            return dp[s];
        }
        auto it = lower_bound(startTime.begin(), startTime.end(), j[s].end);
        
        int index = it - startTime.begin();
        
        dp[s] = max(j[s].prof + solve(index, j,dp, startTime), solve(s+1, j, dp, startTime));
        return dp[s];
    }
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        vector<obj>j;
        vector<int>dp(startTime.size(), -1);
        if(startTime.size() ==1){
            return profit[0];
        }
        for(int i =0; i < startTime.size(); i++){
            j.push_back({startTime[i], endTime[i], profit[i]});

        }
        sort(j.begin(), j.end(), comp());
        sort(startTime.begin(), startTime.end());
        int i =0;
        for(int i =0; i < j.size(); i++){
            solve(i, j, dp, startTime);
        }
        int ma =0;
        for(int i =0 ;i < dp.size(); i++){
            ma = max(ma, dp[i]);
        }  
        return ma; 
    }

};
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