<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:
        jobs = [(start, end, prof) for start, end, prof in zip(startTime, endTime, profit)]
        jobs.sort() # sort by start time ascending

        max_profits = [0] * len(jobs)

        def findMaxProfit(startIdx: int) -> int:
            if max_profits[startIdx] > 0:
                return max_profits[startIdx]

            if startIdx == len(jobs) - 1:
                max_profits[startIdx] = jobs[-1][2]
                return jobs[-1][2]

            # nextIdx is index of first job with startTime >= curr endTime 
            nextIdx = bisect.bisect_left(jobs, jobs[startIdx][1], key=lambda t: t[0])
            schedule_current = 0
            if nextIdx < len(jobs): 
                schedule_current =  jobs[startIdx][2] + findMaxProfit(nextIdx) # schedule current job so must find next idx
            else: # there are no jobs with startTime after this endTime 
                schedule_current = jobs[startIdx][2]

            
            result = max(
                schedule_current,
                findMaxProfit(startIdx + 1),
            )
            max_profits[startIdx] = result
            return result

        return findMaxProfit(0)




        
        '''
        max_profits = [0] * len(startTime) # start maxProfit if we use jobs up to index i 
        # iterate over jobs. at every step, we either include that job or don't include it

        max_profits[0] = jobs[0][2] # max profit of first job
        scheduled_jobs = [jobs[0]]
        for i in range(1, len(jobs)):
            # at each i, either choose to schedule it or not
            # curr start time is after the last end time so we can "add" it
            if jobs[i][0] >= scheduled_jobs[-1][1]: # check end time of last job 
                max_profits[i] = max_profits[i - 1] + jobs[i][2]
                scheduled_jobs.append(jobs[i])
            # if it conflicts with last schedule job, we have to pop until we find an end time that works
            else:
                removed = 0
                idx = len(scheduled_jobs) - 1
                while idx >= 0 and scheduled_jobs[idx][1] > jobs[i][0]: # find job where end time is less than or equal to curr start time 
                    removed = removed + scheduled_jobs[idx][2]
                    idx = idx - 1
                    
                if removed < jobs[i][2]: # only worth removing the others if this profit is better
                    del scheduled_jobs[idx + 1:]
                    scheduled_jobs.append(jobs[i])
                    max_profits[i] = max_profits[i - 1] - removed + jobs[i][2]
                else:
                    max_profits[i] = max_profits[i - 1]

        
        return max_profits[-1]
        '''



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