<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:
        
        def find_next(index):
            left = index + 1
            right = N - 1
            while left <= right:
                mid = left + (right - left) // 2
                # if jobs[mid][0] >= jobs[index][1]:
                #     return mid
                if jobs[mid][0] >= jobs[index][1]:
                    right = mid - 1
                else:
                    left = mid + 1
            return left

        def max_profit(index):
            if index in memo:
                return memo[index]
            if index >= N:
                return 0

            next_index = find_next(index)
            res = max(max_profit(index+1),
                        max_profit(next_index) + jobs[index][-1])
            memo[index] = res
            return res

            # res = float('-inf')
            # 这个for loop是错误的，for loop里面不是当前index的所有选择，而是包含接下来很多步的选择，不需要for loop.
            # for i in range(index+1, N):
            #     if startTime[i] >= end_time:
            #         opt_1 = profit[i] + max_profit(i, endTime[i])
            #         opt_2 = max_profit(i, end_time)
            #         res = res + max(opt_1, opt_2)
            # return res

        N = len(profit)
        jobs = sorted(zip(startTime, endTime, profit))
        memo = {}
        ans = max_profit(0)
        return ans
        
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