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[11/26] 1235. Maximum Profit in Job Scheduling

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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 startTimeendTime and profit arrays, return the maximum profit you can take such that there are no two 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 * 104
  • 1 <= startTime[i] < endTime[i] <= 109
  • 1 <= profit[i] <= 104
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Runtime: 1631 ms, faster than 30.10% of Python3 online submissions for Maximum Profit in Job Scheduling.
Memory Usage: 27.4 MB, less than 65.43% of Python3 online submissions for Maximum Profit in Job Scheduling.
import bisect
class Solution:
    def jobScheduling(self, startTime: List[int], endTime: List[int], profit: List[int]) -> int:
        jobs = sorted(zip(startTime,endTime, profit), key=lambda x:x[1])
        
        dp = [[0,0]]
        
        for s,e,p in jobs:
            i = bisect.bisect_left(dp, [s+1])-1
            if dp[i][1] + p > dp[-1][1]:
                dp.append([e, dp[i][1]+p])
        
        return dp[-1][1]
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