You want to schedule a list of jobs in
d days. Jobs are dependent (i.e To work on the
i-th job, you have to finish all the jobs
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 in that day.
Given an array of integers
jobDifficulty and an integer
d. The difficulty of the
i-th job is
Return the minimum difficulty of a job schedule. If you cannot find a schedule for the jobs return -1.
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
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.
Input: jobDifficulty = [1,1,1], d = 3 Output: 3 Explanation: The schedule is one job per day. total difficulty will be 3.
Input: jobDifficulty = [7,1,7,1,7,1], d = 3 Output: 15
Input: jobDifficulty = [11,111,22,222,33,333,44,444], d = 6 Output: 843
1 <= jobDifficulty.length <= 300
0 <= jobDifficulty[i] <= 1000
1 <= d <= 10