<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>

import math
from collections import namedtuple
import bisect

Job = namedtuple("Job", ["start_time", "end_time", "profit"])

class Solution:
    def jobScheduling(self, startTime: List[int], endTime: List[int], profit: List[int]) -> int:
        print()
        jobs = []
        for i in range(len(startTime)):
            jobs.append(Job(startTime[i], endTime[i], profit[i]))
        jobs.sort()
        # print(jobs)
        
        # map from job index in jobs array to max profit achievable at this start time
        memo_table: dict[int, int] = {} 
        max_profit = -math.inf
        
        for i in range(len(startTime) - 1, -1, -1):
            curr_job = jobs[i]
            print("curr job:", curr_job)
            # curr_profit = curr_job.profit
            # for job_index in memo_table:
            #     if jobs[job_index].start_time >= curr_job.end_time:
            #         curr_profit = max(curr_profit, curr_job.profit + memo_table[job_index])
            profit_job_starts_after_curr = memo_table.get(bisect.bisect_left(jobs, curr_job.end_time, key=lambda x: x.start_time), -math.inf)
            profit_job_one_after_curr =  memo_table.get(i + 1, -math.inf)
            curr_profit = max(curr_job.profit, profit_job_one_after_curr, curr_job.profit + profit_job_starts_after_curr)
            
            max_profit = max(max_profit, curr_profit)
            
            memo_table[i] = max_profit
        
        print("memo_table:", memo_table)
        print("max profit:", max_profit)
        return max_profit
35466