<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 {
    vector<int> dp;
    int solve(vector<pair<int, pair<int, int>>> &vec, int i)
    {
        if(i >= vec.size())
            return 0;
        
        if(dp[i] != -1)
            return dp[i];
        
        int st = vec[i].first;
        int ed = vec[i].second.first;
        int gain = vec[i].second.second;

        int nottake = solve(vec, i+1);
        int take = 0;
        // for(int j=i+1; j<vec.size(); j++)
        // {
        //     if(vec[j].first >= ed){
        //         take = max(take, solve(vec, j));
        //         break;
        //     }
        // }
        auto nextInd = lower_bound(vec.begin(), vec.end(), ed, 
        [](const pair<int, pair<int, int>>& element, int value) {
            return element.first < value;  
        });
        take += gain + solve(vec, nextInd-vec.begin());

        return dp[i] = max(take, nottake);
    }
public:
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        int n = startTime.size();
        vector<pair<int, pair<int, int>>> vec(n);
        dp.resize(n, -1);

        for(int i=0; i<n; i++)
        {
            int st = startTime[i];
            int ed = endTime[i];
            int gain = profit[i];

            vec[i] = {st, {ed, gain}};
        }

        sort(vec.begin(), vec.end());
        return solve(vec, 0);
    }
};
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