<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>

from dataclasses import dataclass

class Solution:
    def bs(self, tasks, startTime):
        if len(tasks) == 0:
            return -1
        n = len(tasks)
        return bisect.bisect_left(tasks, (startTime, startTime, 0))


    def jobScheduling_0(self, startTime: List[int], endTime: List[int], profit: List[int]) -> int:
        tasks = []
        max_et = -1
        for i in range(len(startTime)):
            tasks.append((startTime[i], endTime[i], profit[i]))
        tasks.sort()

        n = len(tasks)
        DP = {}
        def sub(i):
            nonlocal DP, n
            if i > n-1: return 0
            if i == n-1: return tasks[i][2]
            if i in DP:
                return DP[i]

            # pick ith task
            j = self.bs(tasks, tasks[i][1])
            opt1 = tasks[i][2] + sub(j)

            # do not pick i
            opt2 = sub(i+1)
            DP[i] = max(opt1, opt2)
            return DP[i]

        return sub(0)
            
    def jobScheduling(self, startTime: List[int], endTime: List[int], profit: List[int]) -> int:
        @dataclass
        class Task:
            start: int
            end: int
            profit: int

        tasksbyend = defaultdict(list)
        for i,j,p in zip(startTime, endTime, profit):
            t = Task(i,j,p)
            tasksbyend[j].append(t)
       
        endTime = sorted(endTime)
        def prv(end):
            nonlocal endTime
            i = bisect_left(endTime, end)
            return endTime[i-1] if i>0 else None
            
        def round(end):
            nonlocal endTime
            i = bisect_left(endTime, end)
            if endTime[i] == end:
                return end
            elif i>0:
                return endTime[i-1]
            else:
                return 0

        @cache
        def scheduling(end):
            nonlocal tasksbyend
            tasks = tasksbyend[end]
            ans = 0
            for t in tasks:
                ans = max(ans, scheduling(round(t.start))+t.profit)
            if prv(end) is not None:
                ans = max(ans, scheduling(prv(end)))
            #print(end, ans)
            return ans
        
        return scheduling(endTime[-1])
        


            






        
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