1335. 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 i-th 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 in that day.

Given an array of integers jobDifficulty and an integer d. The difficulty of the i-th 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.

Example 4:

Input: jobDifficulty = [7,1,7,1,7,1], d = 3
Output: 15

Example 5:

Input: jobDifficulty = [11,111,22,222,33,333,44,444], d = 6
Output: 843

 

Constraints:

• 1 <= jobDifficulty.length <= 300
• 0 <= jobDifficulty[i] <= 1000
• 1 <= d <= 10

思路：记忆划搜索

class Solution(object):
    def minDifficulty(self, jobDifficulty, d):
        """
        :type jobDifficulty: List[int]
        :type d: int
        :rtype: int
        """
        if len(jobDifficulty)<d: return -1
        if len(jobDifficulty)==d: return sum(jobDifficulty)
        n = len(jobDifficulty)
        INF = 999999999

        memo = {}
        def helper(idx, nd):
            if nd==1: return max(jobDifficulty[idx:])
            if (idx,nd) in memo: return memo[(idx,nd)]
            ma = -1
            res = INF
            for i in range(idx, n-nd+1):
                ma = max(ma, jobDifficulty[i])
                res = min(res, ma+helper(i+1, nd-1))
            memo[(idx,nd)] = res
            return res

        return helper(0, d)

s=Solution()
print(s.minDifficulty(jobDifficulty = [6,5,4,3,2,1], d = 2))
print(s.minDifficulty(jobDifficulty = [7,1,7,1,7,1], d = 3))