<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:
    typedef pair<pair<int, int>, int> P;
    int memo[50001];
    int n;

    int solve(vector<vector<int>>& array, int i) {
        if (i >= n)
            return 0;

        if (memo[i] != -1)
            return memo[i];

        auto it = lower_bound(array.begin() + i + 1, array.end(), vector<int>{array[i][1]}, [](const vector<int>& job, const vector<int>& value) {
                return job[0] < value[0];
            });

        int next = distance(array.begin(), it); 

        int taken = array[i][2] + solve(array, next);
        int notTaken = solve(array, i + 1);

        return memo[i] = max(taken, notTaken);
    }

    int jobScheduling(vector<int>& startTime, vector<int>& endTime,
                      vector<int>& profit) {
        n = startTime.size();

        memset(memo, -1, sizeof(memo));

        vector<vector<int>> array(n, vector<int>(3, 0)); 

        for (int i = 0; i < n; i++) {
            array[i][0] = startTime[i];
            array[i][1] = endTime[i];
            array[i][2] = profit[i];
        }

        sort(begin(array), end(array));

        return solve(array, 0);
    }
};
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