<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:
        ## TLE
        # jobs = sorted(zip(startTime, endTime, profit), key=lambda x: (x[0], x[1], -x[2]))
        # def dp(start_time):
        #     if start_time >= jobs[-1][1]:
        #         return 0
        #     if start_time not in memo:
        #         available_jobs = [j for j in jobs if j[0] >= start_time]
        #         ans = 0
        #         for job in available_jobs:
        #             ans = max(ans, job[2] + dp(job[1]))
        #         memo[start_time] = ans
        #     return memo[start_time]
        # memo = dict()
        # return dp(jobs[0][0])

        ## TLE
        # jobs = sorted(zip(startTime, endTime, profit), key=lambda x: (x[0], x[1], -x[2]))
        # def dp(start_time):
        #     if start_time >= jobs[-1][1]:
        #         return 0
        #     if start_time not in memo:
        #         available_jobs = [j for j in jobs if j[0] >= start_time]
        #         if len(available_jobs) > 0:
        #             first = available_jobs[0]
        #             ans = 0
        #             for job in [j for j in available_jobs if j[0] <= first[1]]:
        #                 ans = max(ans, job[2] + dp(job[1])) 
        #             memo[start_time] = max(ans, dp(first[1]))
        #         else:
        #             memo[start_time] = 0
        #     return memo[start_time]
        # memo = dict()
        # return dp(jobs[0][0])

        jobs = sorted(zip(startTime, endTime, profit), key=lambda x: (x[0], x[1], -x[2]))
        n = len(jobs)
        def dp(start_index):
            if start_index >= n or jobs[start_index][0] >= jobs[-1][1]:
                return 0
            if start_index not in memo:
                # find first available job after start_index
                left = start_index + 1
                right = n - 1
                while left <= right:
                    mid = left + (right - left) // 2
                    if jobs[mid][0] >= jobs[start_index][1]:
                        right = mid - 1
                    else:
                        left = mid + 1
                schedule = jobs[start_index][2] + dp(left)
                notschedule = dp(start_index + 1)
                memo[start_index] = max(schedule, notschedule)
            return memo[start_index]
        memo = dict()
        return dp(0)
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