human
Can you provide a Python program to find a schedule for the jobs return -1 given the following:
Solve in Python:
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
assistant
#Here is Python program to find a schedule for the jobs return -1.

class Solution:
    def minDifficulty(self, jobDifficulty: List[int], d: int) -> int:
        jD=jobDifficulty
        day=d
        self.memo = {}
        def dp(i,day):
            if (i,day) not in self.memo:
                if i < day: self.memo[(i,day)] = float('inf')
                elif day == 1: self.memo[(i,day)] = max(jD[:i])
                elif i == day: self.memo[(i,day)] = sum(jD[:i])
                else:
                    tmp = jD[i-1]
                    self.memo[(i,day)] = dp(i-1,day-1) + tmp
                    for j in range(i-1,day-3,-1):
                        tmp = max(jD[j-1],tmp)
                        self.memo[(i,day)] = min(self.memo[(i,day)],dp(j,day-1) + tmp)
            return self.memo[(i,day)]
        ans = dp(len(jD),d)
        return ans if ans!= float('inf') else -1