<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:
        '''
        Notes

        Approach:
        - use dfs. Either take the current job or not and recursive on that. 
        - need to sort based on start time and group startTime, endTime, and profit

        Example:
        startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]

        - take (1,3,50)
            (cannot take (2,4,10 anymore))
            - take (3,5,40)
                - no more job, update result 90
            - not take (3,5,40)
                - take (3,6,70)
                    - no more job, update result 120
                - not take (3,6,70)
                    - no more job update result 0
        - not take (1,3,50)
            - take (2,4,10)
                - no more job, update result 10
            - not take (2,4,10)
                - take (3,5,40)
                    - no more job, update result 40
                - not take (3,5,40)
                    - take (3,6,70)
                        - no more job, update result 70
                    - not take (3,6,70)
                        - no more job update result 0

        it seems there are redundancy that can be cached. 

        Complexity:
        length of job array
        - Time: O(nlogn) need to sort job array
        - Space: O(n) for graph traversal
        '''

        def dfs(i):
            if i == len(jobs):
                return 0
            if i in cache:
                return cache[i]

            # do not include the current job
            result = dfs(i+1)

            # include the current job

            # need to do binary search instead of linear
            # the commented line below is the linear approach
            #j = i + 1
            #while j < len(jobs):
            #    # find the first instance of job that has beginning time after end time of previous job
            #    if jobs[i][1] <= jobs[j][0]:
            #        break
            #    j += 1
            
            left = 0
            right = len(jobs) - 1
            while left <= right:
                mid = (left + right) // 2
                # new job's start time needs to be equal or greater than prev job end time
                if jobs[i][1] <= jobs[mid][0]:
                    right = mid - 1
                else:
                    left = mid + 1
            j = left

            # library approach
            #j2 = bisect.bisect_right(jobs, (jobs[i][1] , -1, -1))

            cache[i] = max(result, jobs[i][2] + dfs(j))

            return cache[i]

        jobs=[]
        for i in range(len(startTime)):
            jobs.append((startTime[i], endTime[i], profit[i]))
        
        #jobs.sort(key=lambda x: x[0])
        jobs.sort()
        cache = {} 
        return dfs(0)

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