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
class Solution:
    def minDifficulty(self, jobDifficulty: List[int], d: int) -> int:
        n, inf = len(jobDifficulty), float('inf')
        
        dp = [[inf] * n + [0] for _ in range(d + 1)]
        
        for d in range(1, d + 1):
            for i in range(n - (d - 1)):
                maxd = 0
                for j in range(i, n - (d - 1)):
                    maxd = max(maxd, jobDifficulty[j])
                    dp[d][i] = min(dp[d][i], maxd + dp[d - 1][j + 1])
        print(dp)
        return dp[d][0] if dp[d][0] < inf else -1
class Solution:
    def minDifficulty(self, jobDifficulty: List[int], d: int) -> int:
        n, inf = len(jobDifficulty), float('inf')
        
        dp = [[inf] * n + [0] for _ in range(d + 1)]
        
        for d in range(1, d + 1):
            for i in range(n - (d - 1)):
                maxd = 0
                for j in range(i, n - (d - 1)):
                    maxd = max(maxd, jobDifficulty[j])
                    dp[d][i] = min(dp[d][i], maxd + dp[d - 1][j + 1])
        print(dp)
        return dp[d][0] if dp[d][0] < inf else -1