<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:
    class node
    {
        public:
        int start;
        int end;
        int profit;

        node(int _s,int _e,int _p)
        {
            start = _s;
            end = _e;
            profit = _p;
        }
    };

    static bool cmp(node* a,node* b)
    {
        if(a->start < b->start)
        return true;

        return false;
    }

    int solve(vector<node*>&v,int ind,vector<int>&start,vector<int>&dp)
    {
        if(ind >= v.size())
        return 0;

        if(dp[ind] != -1)
        return dp[ind];

        int j = lower_bound(start.begin()+(ind+1),start.end(),v[ind]->end) - start.begin();

        int nt = 0 + solve(v,ind+1,start,dp);

        int t = v[ind]->profit + solve(v,j,start,dp);

        return dp[ind] = max(nt,t);
    }

    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        vector<node*> v;

        for(int i=0;i<profit.size();i++)
        {
            v.push_back(new node(startTime[i],endTime[i],profit[i]));
        }

        sort(v.begin(),v.end(),cmp);

        for(int i=0;i<v.size();i++)
        startTime[i] = v[i]->start;

        vector<int> dp(v.size()+1,-1);
        int ans = solve(v,0,startTime,dp);
        return ans;
    }
};
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