<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>

struct Job {
  int startTime;
  int endTime;
  int profit;
};

class Solution {
 public:
  int jobScheduling(vector<int>& startTime, vector<int>& endTime,
                    vector<int>& profit) {
    const int n = startTime.size();
    vector<int> mem(n + 1);
    vector<Job> jobs;

    for (int i = 0; i < n; ++i)
      jobs.emplace_back(startTime[i], endTime[i], profit[i]);

    ranges::sort(jobs, ranges::less{},
                 [](const Job& job) { return job.startTime; });

    // Will use binary search to find the first available start time.
    for (int i = 0; i < n; ++i)
      startTime[i] = jobs[i].startTime;

    return jobScheduling(jobs, startTime, 0, mem);
  }

 private:
  // Returns the maximum profit to schedule jobs[i..n).
  int jobScheduling(const vector<Job>& jobs, const vector<int>& startTime,
                    int i, vector<int>& mem) {
    if (i == jobs.size())
      return 0;
    if (mem[i] > 0)
      return mem[i];

    const int j = firstGreaterEqual(startTime, i + 1, jobs[i].endTime);
    const int pick = jobs[i].profit + jobScheduling(jobs, startTime, j, mem);
    const int skip = jobScheduling(jobs, startTime, i + 1, mem);
    return mem[i] = max(pick, skip);
  }

  int firstGreaterEqual(const vector<int>& A, int startFrom, int target) {
    return lower_bound(A.begin() + startFrom, A.end(), target) - A.begin();
  }
};

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