<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:
        if len(profit) <= 1:
            return profit[0]
        
        maxt = max(endTime)
        ps = []
        h = []
        for i in range(len(endTime)):
            heapq.heappush(h, (endTime[i], startTime[i], profit[i]))
        while len(h) > 0:
            end,start,p = heapq.heappop(h)
            if len(ps) == 0:
                ps.append((end,p))
            elif ps[-1][0] < end:
                lastP = ps[-1][1]
                curP = max(self.find(ps,start) + p, lastP)
                ps.append((end,curP))
            else:
                sameTimeP = ps[-1][1]
                lastP = ps[-2][1] if len(ps)>=2 else 0
                curP = max(self.find(ps,start) + p, sameTimeP, lastP)
                ps.pop()
                ps.append((end,curP))
        return ps[-1][1]
    
    def find(self, ps, start):
        if start < ps[0][0]:
            return 0
        l = 0
        r = len(ps)-1
        while l<r:
            mid = (l+r+1)//2
            if ps[mid][0] == start:
                return ps[mid][1]
            if ps[mid][0] < start:
                l = mid
            else:
                r = mid-1
        return ps[r][1]
        

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