<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:

    struct node {
        int st,et,p;
        node(int a, int b, int c) {
            st=a,et=b,p=c;
        }
    };

    unordered_map<int,int> dp;

    int getNext(vector<node*> &v, int x) {
        int l=0,r=v.size()-1;
        while(l!=r) {
            int md=(l+r)/2;
            if(v[md]->st >= x) {
                r=md;
            } else {
                l=md;
            }
            if(l+1==r) {
                if(v[l]->st >= x) r=l;
                else l=r;
                break;
            }
        }
        if(v[l]->st >= x) return l;
        return -1;
    }

    int deepee(vector<node*> &v, int x) {
        if(dp[x]) return dp[x];
        int mx=v[x]->p;
        int idx=getNext(v,v[x]->et);
        if(idx!=-1) {
            mx+=deepee(v,idx);
        }
        if(x+1!=v.size()) {
            mx=max(mx,deepee(v,x+1));
        }
        dp[x]=mx;
        return mx;
    }

    static bool sorter(node *l, node *r) {
        return l->st < r->st;
    }

    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        dp.clear();
        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(),sorter);
        return deepee(v,0);
    }
};
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