<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 {
    int maximize(vector<pair<pair<int,int>,int>>& okay, int index, int n, vector<int>& memo) {
        if (index >= n) {
            return 0;
        }
        
        if (memo[index] != -1) {
            return memo[index];
        }
        
        // Option 1: Skip the current job
        int skipCurrent = maximize(okay, index + 1, n, memo);
        
        // Option 2: Include the current job
        int includeCurrent = okay[index].second;
        
        // Find the next job that can be included after this one
        int nextIndex = index + 1;
        while (nextIndex < n && okay[nextIndex].first.first < okay[index].first.second) {
            nextIndex++;
        }
        
        if (nextIndex < n) {
            includeCurrent += maximize(okay, nextIndex, n, memo);
        }
        
        // Store the result in memo array
        memo[index] = max(includeCurrent, skipCurrent);
        return memo[index];
    }

public:
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        vector<pair<pair<int,int>,int>> okay;
        int n = startTime.size();
        vector<int> memo(n, -1);
        
        for (int i = 0; i < n; i++) {
            okay.push_back({{startTime[i], endTime[i]}, profit[i]});
        }
        
        // Sort jobs by their start times
        sort(okay.begin(), okay.end(), [](pair<pair<int,int>,int>& a, pair<pair<int,int>,int>& b) {
            return a.first.first < b.first.first;
        });
        
        return maximize(okay, 0, n, memo);
    }
};

85133