Class Solution


  • public class Solution
    extends Object
    1335 - Minimum Difficulty of a Job Schedule\. Hard You want to schedule a list of jobs in `d` days. Jobs are dependent (i.e To work on the ith job, you have to finish all the jobs `j` where `0 <= j < i`). You have to finish **at least** one task every day. The difficulty of a job schedule is the sum of difficulties of each day of the `d` days. The difficulty of a day is the maximum difficulty of a job done on that day. You are given an integer array `jobDifficulty` and an integer `d`. The difficulty of the ith job is `jobDifficulty[i]`. Return _the minimum difficulty of a job schedule_. If you cannot find a schedule for the jobs return `-1`. **Example 1:** ![](https://assets.leetcode.com/uploads/2020/01/16/untitled.png) **Input:** jobDifficulty = [6,5,4,3,2,1], d = 2 **Output:** 7 **Explanation:** First day you can finish the first 5 jobs, total difficulty = 6. Second day you can finish the last job, total difficulty = 1. The difficulty of the schedule = 6 + 1 = 7 **Example 2:** **Input:** jobDifficulty = [9,9,9], d = 4 **Output:** -1 **Explanation:** If you finish a job per day you will still have a free day. you cannot find a schedule for the given jobs. **Example 3:** **Input:** jobDifficulty = [1,1,1], d = 3 **Output:** 3 **Explanation:** The schedule is one job per day. total difficulty will be 3. **Constraints:** * `1 <= jobDifficulty.length <= 300` * `0 <= jobDifficulty[i] <= 1000` * `1 <= d <= 10`
    • Constructor Detail

      • Solution

        public Solution()
    • Method Detail

      • minDifficulty

        public int minDifficulty​(int[] jobDifficulty,
                                 int d)