<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:
int maxProfitUtil(int index, const vector<vector<int>>& jobs, map<int, int>& memo) {
    // Base case: if we've processed all jobs, return 0 profit
    if (index >= jobs.size()) return 0;

    // If we've already solved this subproblem, return the result
    if (memo.find(index) != memo.end()) return memo[index];

    // Option 1: Skip the current job
    int skipProfit = maxProfitUtil(index + 1, jobs, memo);

    // Option 2: Take the current job and find the next job that starts after this job ends
    int nextJobIndex = lower_bound(jobs.begin() + index + 1, jobs.end(), jobs[index][1],
                                   [](const vector<int>& job, int endTime) {
                                       return job[0] < endTime;
                                   }) - jobs.begin();
    int takeProfit = jobs[index][2] + maxProfitUtil(nextJobIndex, jobs, memo);

    // Store the maximum profit for this subproblem
    memo[index] = max(skipProfit, takeProfit);

    return memo[index];
}

public:
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
     int n = startTime.size();
    vector<vector<int>> jobs(n, vector<int>(3));

    // Create jobs array where each job is a vector of [startTime, endTime, profit]
    for (int i = 0; i < n; ++i) {
        jobs[i] = {startTime[i], endTime[i], profit[i]};
    }

    // Sort jobs by start time
    sort(jobs.begin(), jobs.end(), [](const vector<int>& a, const vector<int>& b) {
        return a[0] < b[0];
    });

    // Map to store the results of subproblems
    map<int, int> memo;

    // Calculate the maximum profit starting from the first job
    return maxProfitUtil(0, jobs, memo);
    }
};
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