<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 {
private:
    static const int maxn = 50000;
    int indices[maxn + 10];
    int n;
    int p[maxn];
    int seg[4 * maxn + 1];
    inline void update(int x, int value, int p = 0 , int l = 0 , int r = maxn - 1){
        if (l == r){
            seg[p] = max(seg[p], value);
            return;
        }
        int mid = (l + r) >> 1;

        if (x < mid){
            update(x, value, 2 * p + 1, l, mid);
        } else {
            update(x, value, 2 * p + 2, mid + 1, r);
        }

        seg[p] = max(seg[2 * p + 1], seg[2 * p + 2]);
    }
    inline int get_max(int x, int p = 0 , int l = 0 , int r = maxn - 1){
        if (l > r){
            return 0;
        }
        if (l == r){
            return seg[p];
        } 

        int mid = (l + r) >> 1;
        if (x < mid){
            return get_max(x, 2 * p + 1 , l , mid);
        } else {
            return max(seg[2 * p + 1], get_max(x, 2 * p + 2, mid + 1, r));
        }
    }
public:
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {

        n = startTime.size();
    
        iota(indices, indices + n, 0);

        sort(indices, indices + n, [&endTime](const int&a, const int&b) {
            return endTime[a] < endTime[b];
        });

        int response = 0;
        for (int i = 0 ; i < n ; i ++){
            int low = startTime[indices[i]];

            int l_search = 0;
            int h_search = i;
            while (l_search + 1 < h_search){
                int mid = (l_search + h_search) >> 1;
                if (endTime[indices[mid]] <= low){
                    l_search = mid;
                } else {
                    h_search = mid;
                }
            }


            int best = 0;
            
            if (h_search > 0 && endTime[indices[h_search - 1]] <= low){
                best = profit[indices[i]] + get_max(h_search - 1);
            } else {
                best = profit[indices[i]];
            }
            
            update(i, best);

            // if (best == 11){
            //     cout << "hey\n";
            //     cout <<  (h_search) << "\n";
            //     cout << i << "\n";
            // }

            response = max(response, best);
        }

        return response;
    }
};
73230