java.lang.Object
g1201_1300.s1235_maximum_profit_in_job_scheduling.Solution

public class Solution extends Object
1235 - Maximum Profit in Job Scheduling\. Hard 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, 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:** **![](https://assets.leetcode.com/uploads/2019/10/10/sample1_1584.png)** **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:** **![](https://assets.leetcode.com/uploads/2019/10/10/sample22_1584.png)** **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:** **![](https://assets.leetcode.com/uploads/2019/10/10/sample3_1584.png)** **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
  • Constructor Details

    • Solution

      public Solution()
  • Method Details

    • jobScheduling

      public int jobScheduling(int[] startTime, int[] endTime, int[] profit)