## # 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
``````

Approach 1: DP.

### # Code (Python)

Approach 1:

``````class Solution:
def minDifficulty(self, difficulty: List[int], days: int) -> int:
'''
# TLE solution
# value[i][j][k]: min difficulty to do job[i:j+1] in k days, 1 <= k <= d
# value[i][j] = max(difficulty[i:j+1])
# value[i][j][k] = min(value[i][m] + value[m+1][j][k-1]) for i <= m <= j - k
if d > len(difficulty):
return -1
if d == len(difficulty):
return sum(difficulty)
if d == 1:
return max(difficulty)

value = [[[float('inf')] * len(difficulty) for _ in range(len(difficulty))] for _ in range(d + 1)]
for i in range(len(difficulty)):
for j in range(i, len(difficulty)):
value[i][j] = max(difficulty[i:j + 1])
for k in range(2, d + 1):
for i in range(len(difficulty)):
for j in range(i, len(difficulty)):
for m in range(i, j - k + 2):
value[k][i][j] = min(value[k][i][j], value[i][m] + value[k - 1][m + 1][j])
return value[-1][-1]
'''
# value[i][d]: min difficulty to do job[i:] in d days, need to return value[d]
# value[i][d] = min(max(difficulty[i:k]) + value[k][d-1]) for k in range(i + 1, ...)
value = [[float('inf')] * len(difficulty) for _ in range(days + 1)]
for d in range(1, days + 1):
for i in range(len(difficulty) - 1, -1, -1):
if d == 1:
value[d][i] = max(difficulty[i:])
else:
max_so_far = -float('inf')
for k in range(i, len(difficulty)):
max_so_far = max(max_so_far, difficulty[k])
if k + 1 < len(difficulty):
value[d][i] = min(value[d][i], max_so_far + value[d-1][k+1])

return value[d] if value[d] != float('inf') else -1
``````

Approach 1:

Approach 2: