<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:
        
        # non-overlapping means sort by the start times
        # if endtime1 == starttime2 you can schedule
        # also want maximum

        # recurrence relation: max possible profit if you choose a particular job
        # loop through all jobs
        # maintain a set of chosen jobs
        # and at each step, after choosing a job, the set of doable jobs changes

        # should return max profit
        # state should represent which jobs have been chosen so far (the sequence of jobs chosen)

        zipped_lists = zip(startTime, endTime, profit)
        sorted_lists = sorted(zipped_lists)
        startTime, endTime, profit = zip(*sorted_lists)


        def no_overlap(i, j):
            nonlocal startTime
            nonlocal endTime

            if endTime[i] <= startTime[j]:
                return True

            else:
                return False 


        # [0,1,2,2,3,3,5]
        # return the first occurrence of a non-overlapping index
        def b_search(i, j):
            # two ends of the array
            left = i
            right = j
            res = -1

            while left < right:
                mid = (left + right) // 2

                if no_overlap(i, mid):
                    right = mid 
                    res = 1

                else:
                    left = mid + 1

            if res == 1:
                return left

            else:
                return -1



        # assume the arrays are sorted
        max_profits = [0] * len(startTime)
        max_profits[-1] = profit[-1]

        def dp():
            # maximum is = current, current + 1, or current + first non-overlap
            nonlocal max_profits
            for i in range(len(max_profits)-2, -1, -1):
                best_profit = max(profit[i], max_profits[i+1])
                j = b_search(i, len(max_profits))

                if j!=-1:
                    best_profit = max(best_profit, profit[i]+max_profits[j])

                # for j in range(i+1, len(max_profits)):

                #     # we care about the first non-overlap
                #     if no_overlap(i,j):
                #         # see if this makes things better or not
                #         best_profit = max(best_profit, profit[i]+max_profits[j])
                #         break

                max_profits[i] = best_profit
            #print(max_profits)
            return max_profits[0]

        return dp()



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