LeetCode — Minimum Difficulty of a Job Schedule
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
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
Return the minimum difficulty of a job schedule. If you cannot find a schedule for the jobs return
Input: jobDifficulty = [6,5,4,3,2,1], d = 2
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
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
Explanation: The schedule is one job per day. total difficulty will be 3.
1 <= jobDifficulty.length <= 300
0 <= jobDifficulty[i] <= 1000
1 <= d <= 10
Time complexity: O(n² * k)
Space complexity: O(n)