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1335. Minimum Difficulty of a Job Schedule

Difficulty Topics

Description

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:

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

Solution

minimum-difficulty-of-a-job-schedule.py
class Solution:
    def minDifficulty(self, jobs: List[int], d: int) -> int:
        N = len(jobs)
        if N < d: return -1

        @cache
        def go(index, day):
            if day == d: return 0 if index == N else inf

            res = inf
            curr = 0
            for j in range(index, N):
                curr = max(curr, jobs[j])
                res = min(res, curr + go(j + 1, day + 1))

            return res

        return go(0, 0)