Class Solution
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public final class Solution1235 - Maximum Profit in Job Scheduling.
Hard
We have
njobs, where every job is scheduled to be done fromstartTime[i]toendTime[i], obtaining a profit ofprofit[i].You're given the
startTime,endTimeandprofitarrays, 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
Xyou will be able to start another job that starts at timeX.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:
<code>1 <= startTime.length == endTime.length == profit.length <= 5 * 10<sup>4</sup></code>
<code>1 <= startTimei< endTimei<= 10<sup>9</sup></code>
<code>1 <= profiti<= 10<sup>4</sup></code>
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Constructor Summary
Constructors Constructor Description Solution()
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Method Summary
Modifier and Type Method Description final IntegerjobScheduling(IntArray startTime, IntArray endTime, IntArray profit)-
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Method Detail
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jobScheduling
final Integer jobScheduling(IntArray startTime, IntArray endTime, IntArray profit)
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