Minimum difficulty of a job schedule

Time: O(DxN^2); Space: O(DxN); hard

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 <= len(jobDifficulty) <= 300

  • 0 <= jobDifficulty[i] <= 1000

  • 1 <= d <= 10

Hints:

  1. Use DP. Try to cut the array into d non-empty sub-arrays. Try all possible cuts for the array.

  2. Use dp[i][j] where DP states are i the index of the last cut and j the number of remaining cuts. Complexity is O(n * n * d).

1. Dynamic programming [O(DxN^2), O(DxN)]

[1]:

class Solution1(object):
    """
    Time: O(D*N^2)
    Space: O(D*N)
    """
    def minDifficulty(self, jobDifficulty, d):
        """
        :type jobDifficulty: List[int]
        :type d: int
        :rtype: int
        """
        if len(jobDifficulty) < d:
            return -1;

        dp = [[float("inf")]*len(jobDifficulty) for _ in range(d)]
        dp[0][0] = jobDifficulty[0]

        for i in range(1, len(jobDifficulty)):
            dp[0][i] = max(dp[0][i-1], jobDifficulty[i])

        for i in range(1, d):
            for j in range(i, len(jobDifficulty)):
                curr_max = jobDifficulty[j]
                for k in reversed(range(i, j+1)):
                    curr_max = max(curr_max, jobDifficulty[k])
                    dp[i][j] = min(dp[i][j], dp[i-1][k-1] + curr_max)

        return dp[d-1][len(jobDifficulty)-1]

[2]:

s = Solution1()

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

jobDifficulty = [9,9,9]
d = 4
assert s.minDifficulty(jobDifficulty, d) == -1

jobDifficulty = [1,1,1]
d = 3
assert s.minDifficulty(jobDifficulty, d) == 3

jobDifficulty = [7,1,7,1,7,1]
d = 3
assert s.minDifficulty(jobDifficulty, d) == 15

jobDifficulty = [11,111,22,222,33,333,44,444]
d = 6
assert s.minDifficulty(jobDifficulty, d) == 843