1335. Minimum Difficulty of a Job Schedule

Hard

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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

class Solution:
    def minDifficulty(self, jobDifficulty: List[int], d: int) -> int:
        length = len(jobDifficulty)
        if length < d:
            return -1
        @lru_cache(None)
        def dfs(index, d):
            if d == 1:
                return max(jobDifficulty[index:])
            result, maxd = float("inf"), 0
            for i in range(index, length-d+1):
                maxd = max(maxd, jobDifficulty[i])
                result = min(result, maxd + dfs(i+1, d-1))
            return result
        return dfs(0, d)