<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:
        
        max_profit = 0

        n_jobs = len(startTime)

        # sort based on startTime
        sorted_arrays = sorted(zip(startTime, endTime, profit))
        sorted_st, sorted_en, sorted_pr = zip(*sorted_arrays)

        print(sorted_st)

        @lru_cache(maxsize=None)
        def recurse(list_position):
            if list_position > n_jobs - 1:
                return 0
            
            current_job_profit = sorted_pr[list_position]
            end_time_if_scheduled = sorted_en[list_position]
            #print('Finding next position from list_pos {:d}'.format(list_position))
            next_position_if_scheduled = find_new_position(list_position, end_time_if_scheduled)
            #print('Next position is {:d}'.format(next_position_if_scheduled))

            p1 = recurse(list_position + 1)
            p2 = recurse(next_position_if_scheduled) + current_job_profit

            #print('Lpos: {:d}, Lpos++: {:d}, ET SCH: {:d}, NP SCH: {:d}, MaxPr: {:d}'.format(list_position, list_position+1, end_time_if_scheduled, next_position_if_scheduled, max(p1, p2)))
            return max(p1, p2)

        # conduct a binary search to find the next target position
        def find_new_position(list_position, end_time):
            l = list_position + 1
            r = n_jobs-1
            pos = n_jobs

            while l <= r:
                
                mid = l + (r-l)//2
                #print(l, mid, r, end_time, pos)
                if sorted_st[mid] >= end_time:
                    pos = mid
                    #print('pos/mid set to {:d} with time value {:d}'.format(pos, sorted_en[mid]))
                    r = mid - 1
                else:
                    l = mid + 1
                
                if sorted_st[r] < end_time:
                    #print('Special exit', pos)
                    return pos

            print(pos)
            return pos

        total_profit = recurse(0)

        return total_profit

        
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