<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:
        tbl = []
        for x, y, z in zip(startTime, endTime, profit):
            # we need to sort first by endTime, then by startTime
            # so we put endTime first, then it is automatically
            # the first sorting key
            tbl.append([y, x, z])
        
        # sort first by endTime, then by startTime
        tbl.sort()
        # we need a sorted list of endTime to find next job to deal with
        endTime.sort() 
        # print(tbl)


        # input: idx - index of job in the sorted tbl 
        # output: maxium profit from index 0 to index idx 
        @cache
        def dp(idx):
            if idx < 0:
                # in this case, all jobs were already considered
                # no more profit, so we return 0
                return 0 
            
            # we do not choose job No.idx, 
            # then just consider the previous job
            # ans1 is the profit we get if current job is not added
            ans1 = dp(idx-1)

            # print(tbl[idx][1])
            # we choose job No.idx 
            # calculate which job should be considered next 
            # nxt is the index of the next job to be considered 
            nxt = bisect_left(endTime, tbl[idx][1])
            if tbl[nxt][0] > tbl[idx][1]:
                nxt -= 1
            elif tbl[nxt][0] == tbl[idx][1]:
                nxt = bisect_right(endTime, tbl[idx][1]) - 1
            # print(nxt, tbl[nxt][0],tbl[idx][1])
            # print(nxt)
            
            # ans2 is the profit we get if current job is added
            ans2 = dp(nxt) + tbl[idx][2]
            # print("idx: ", idx, ans1, ans2)

            # compare and decide the bigger profit 
            return max(ans1, ans2)
        
        return dp(len(tbl)-1)
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