<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>

# https://www.youtube.com/watch?v=JLoWc3v0SiE
class Solution:
    def jobScheduling(self, startTime: List[int], endTime: List[int], profit: List[int]) -> int:
        tasks = sorted(zip(startTime, endTime, profit), key=lambda record: record[0])
        @cache
        def recurse(i):
            '''Returns max profit starting at task i.
            '''
            if i >= len(tasks):
                return 0
            if i == len(tasks) - 1:
                return tasks[i][2]
            next_valid_task = bisect_left(tasks, tasks[i][1], key=lambda record: record[0])
            take_profit = tasks[i][2] + recurse(next_valid_task)
            notake_profit = recurse(i+1)
            return max(take_profit, notake_profit)
        return recurse(0)












        # tasks = sorted(zip(startTime, endTime, profit))
        # # At each task i, I can either take it or not take it.
        # # - if I take it, I reduced the problem to a subset of remaining tasks that don't overlap.
        # # - if I don't take ie, I reduced the problem to tasks[i:].
        # #
        # # Note: taking task as input will lead to higher cache miss, which causes max recursion depth exceeded
        # # error. Let's take task id direclty.
        # @cache
        # def backtrack(task_id):
        #     if task_id >= len(tasks):
        #         return 0
        #     if task_id == len(tasks) - 1:
        #         return tasks[task_id][2]
        #     # Find remaining tasks profit if I take this task:
        #     next_task_id = bisect_left(tasks, (tasks[task_id][1], 0, 0))
        #     take_profit = tasks[task_id][2] + backtrack(next_task_id)
        #     # Find remaining tasks profit if I don't take this task:
        #     notake_profit = backtrack(task_id+1)
        #     return max(take_profit, notake_profit)
        # return backtrack(0)
            
        # def backtrack(task):
        #     # Base case: last task.
        #     if task == tasks[-1]:
        #         return task[2]
        #     # Find remaining tasks if I take this task:
        #     take_task_id = bisect_left(tasks, (task[1], 0, 0))
        #     take_profit = backtrack(tasks[take_task_id]) + task[2] if take_task_id < len(tasks) else task[2]
        #     notake_task_id = bisect_left(tasks, task) + 1
        #     notake_profit = backtrack(tasks[notake_task_id]) if notake_task_id < len(tasks) else task[2]
        #     return max(take_profit, notake_profit)
        # return backtrack(tasks[0])


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