<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>

# endTime -> [(startTime, profit)]
# dp[index_i] tracks the max profit end at index_i, if i job doesn't overlap with previous, 
# https://leetcode.com/problems/maximum-profit-in-job-scheduling/solutions/2882702/python-w-explanation/?envType=problem-list-v2&envId=mb4cupb7
# https://leetcode.com/problems/maximum-profit-in-job-scheduling/solutions/1431246/easiest-heap-98-faster-clean-concise/?envType=problem-list-v2&envId=mb4cupb7
class Solution:
    def jobScheduling(self, startTime: List[int], endTime: List[int], profit: List[int]) -> int:
        # maps = collections.defaultdict(list)
        # for index in range(len(startTime)):
        #     maps[endTime[index]].append((startTime[index], profit[index]))
        # startTime.sort()
        # endTime.sort()
        # times = []
        # i, j = 0, 0
        # while i < len(startTime) or j < len(startTime):
        #     if i >= len(startTime):
        #         times.extend(endTime[j:])
        #         j = len(startTime)
        #     elif j >= len(startTime):
        #         times.extend(startTime[i:])
        #         i = len(startTime)
        #     elif startTime[i] <= endTime[j]:
        #         times.append(startTime[i])
        #         i += 1
        #     else:
        #         times.append(endTime[j])
        #         j += 1
        # startMap = {}
        # dp = [0]*(len(times) + 1)
        # for i in range(1, len(times) + 1):
        #     time = times[i-1]
        #     startMap[time] = i
        #     dp[i] = dp[i-1]
        #     for start, profit in maps[time]:
        #         dp[i] = max(dp[i], dp[startMap[start]] + profit)
        # return max(dp)
        jobs = sorted(zip(endTime, startTime, profit))
        endSeries = [jobs[i][0] for i in range(len(endTime))]
        dp = [0]* (len(startTime) + 1)
        for index, job in enumerate(jobs):
            _, start, profit = job
            startIndex = bisect.bisect_right(endSeries, start)
            dp[index+1] = max(dp[index +1], profit, dp[index], dp[startIndex] + profit)
        return dp[-1]


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