<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:
    struct Job
    {
        int start, end, profit;
    };
    // bool cmp(Job a, Job b)
    // {
    //     return a.end < b.end;
    // }
    int find_last_nonconflicting_job(vector<Job> &jobs, int level)
    {
        int lo = 0;
        int hi = level - 1;
        while (lo <= hi)
        {
            int mid = lo + (hi - lo) / 2;
            if (jobs[mid].end <= jobs[level].start)
            {
                if (jobs[mid + 1].end <= jobs[level].start)
                {
                    lo = mid + 1;
                }
                else
                {
                    return mid;
                }
            }
            else
            {
                hi = mid - 1;
            }
        }
        return -1;
    }
    int dp[50010];
    int rec(int level, vector<Job> &jobs)
    {
        if (level == 0)
        {
            return jobs[0].profit;
        }
        if (dp[level] != -1)
        {
            return dp[level];
        }
        int ans = jobs[level].profit;
        int last_nonconflicting_job = find_last_nonconflicting_job(jobs, level);
        if (last_nonconflicting_job != -1)
        {
            ans += rec(last_nonconflicting_job, jobs);
        }
        ans = max(ans, rec(level - 1, jobs));
        return dp[level] = ans;
    }
    int jobScheduling(vector<int> &startTime, vector<int> &endTime, vector<int> &profit)
    {
        vector<Job> jobs;
        for (int i = 0; i < startTime.size(); i++)
        {
            jobs.push_back({startTime[i], endTime[i], profit[i]});
        }
        sort(jobs.begin(), jobs.end(), [](Job a, Job b)
             { return a.end < b.end; });
        int n = jobs.size();
        memset(dp, -1, sizeof(dp));
        return rec(n - 1, jobs);
    }
};
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