<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:
        n = len(startTime)
        dp = [0] * n
        sort_st_index = [i[0] for i in sorted(enumerate(startTime), key=lambda x:x[1])]
        dp[n - 1] = profit[sort_st_index[n - 1]]
        for i in range(n - 2, -1, -1):
            l = i + 1
            r = n - 1
            while l < r:
                mid = (l + r) // 2
                if endTime[sort_st_index[i]] > startTime[sort_st_index[mid]]:
                    l = mid + 1
                else:
                    r = mid
            sprofit = profit[sort_st_index[i]]
            nsprofit = dp[i + 1]
            if endTime[sort_st_index[i]] <= startTime[sort_st_index[r]]:
                sprofit += dp[r]
            dp[i] = max(sprofit, nsprofit)
        return dp[0]

        '''
        n = len(startTime)
        sort_st_index = [i[0] for i in sorted(enumerate(startTime), key=lambda x:x[1])]
        scache = {}
        nscache = {}
        def cal_profit(ind): # 0, 5, 7
            if ind == n - 1:
                scache[ind] = profit[sort_st_index[ind]]
                nscache[ind] = 0
                return profit[sort_st_index[ind]] # 

            sprofit = profit[sort_st_index[ind]] # 5, 19
            nsprofit = 0
            l = ind + 1
            r = n - 1

            while l < r:
                mid = (l + r) // 2
                if endTime[sort_st_index[ind]] > startTime[sort_st_index[mid]]:
                    l = mid + 1
                else:
                    r = mid
            i = r
            
            if endTime[sort_st_index[ind]] <= startTime[sort_st_index[i]]:
                if i in scache:
                    sprofit += scache[i]
                else:
                    sprofit += cal_profit(i) # 5, 7 | -, 22 
            
            if ind + 1 < n:
                if ind + 1 in scache:
                    nsprofit += scache[ind + 1]
                else:
                    nsprofit = cal_profit(ind + 1) # 70
            
            scache[ind] = max(sprofit, nsprofit) # 3
            return max(sprofit, nsprofit) # 3

        return cal_profit(0)
        '''
748