<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:
    def jobScheduling(self, startTime: List[int], endTime: List[int], profit: List[int]) -> int:
        # sorted by end time
        jobs = sorted(zip(startTime, endTime, profit), key=lambda x:x[1])
        # an array of end times
        endtimes = [e for s,e,p in jobs]
        # dp array returns the most optimal job profit
        dpProfit = [0] * len(jobs)
        dpProfit[0] = jobs[0][2] # initialize with first profit

        # Go through the jobs, there are cases when you don’t schedule, and cases when you do. 
        for i in range(1, len(jobs)):
            start, end, profit = jobs[i][0], jobs[i][1], jobs[i][2]
            # When you don’t: you are saving the next state of dp with the previous state 
            dpProfit[i] = dpProfit[i-1]
            # When you do: Since you are going from lesser end time to greater end time,
            # you want to iterate through the endtimes array and identify the last job that 
            # ends before or at the current job's startTime. 
            # That index would be in the past.
            # If you don't subtract 1, you'd be pointing to a job that overlaps with the 
            # current job because that job ends after the current job starts.
            index = self.bSearch(endtimes, start) - 1
            # So new comparable profit is the dp at that index + curr profit OR the current dp profit
            dpProfit[i] = max(dpProfit[i], profit + (dpProfit[index] if index >= 0 else 0))
        return dpProfit[-1]

    def bSearch(self, nums, n):
        # find the first job that ends after the current job's startTime.
        left, right = 0, len(nums)
        while left + 1 < right:
            mid = left + (right-left)//2
            if nums[mid] > n: # Searching for first endtime greater than start
                right = mid
            else:
                left = mid
        if nums[left] > n:
            return left
        else:
            return right

# bSearch(endtimes, start) gives the index of the first job that ends after the start time of the current job.
# bSearch(endtimes, start) - 1 gives the index of the last job that ends before or at the start time
 
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