human
Solve in Python:
We have n jobs, where every job is scheduled to be done from startTime[i] to endTime[i], obtaining a profit of profit[i].
You're given the startTime , endTime and profit arrays, you need to output the maximum profit you can take such that there are no 2 jobs in the subset with overlapping time range.
If you choose a job that ends at time X you will be able to start another job that starts at time X.
 
Example 1:

Input: startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]
Output: 120
Explanation: The subset chosen is the first and fourth job. 
Time range [1-3]+[3-6] , we get profit of 120 = 50 + 70.

Example 2:
 

Input: startTime = [1,2,3,4,6], endTime = [3,5,10,6,9], profit = [20,20,100,70,60]
Output: 150
Explanation: The subset chosen is the first, fourth and fifth job. 
Profit obtained 150 = 20 + 70 + 60.

Example 3:

Input: startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4]
Output: 6

 
Constraints:

1 <= startTime.length == endTime.length == profit.length <= 5 * 10^4
1 <= startTime[i] < endTime[i] <= 10^9
1 <= profit[i] <= 10^4

Give me a the Python code you requested?
assistant
#Here is the Python code you requested.

class Solution:
    def jobScheduling(self, startTime: List[int], endTime: List[int], profit: List[int]) -> int:
        line=list(set(startTime)|set(endTime))
        line.sort()
        dp=dict()
        d=dict()
        v=dict()
        for i in range(len(endTime)):
            if endTime[i] not in d:
                d[endTime[i]]=[]
            d[endTime[i]].append(startTime[i])
            v[(startTime[i],endTime[i])]=profit[i]
        m=0
        for n in line:
            if n not in d:
                dp[n]=m
            else:
                dp[n]=max([m]+[dp[i]+v[(i,n)]  for i in d[n]])
                m=max(dp[n],m)
        return dp[line[-1]]