# LeetCode — 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 `ith`

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 on that day.

You are given an integer array `jobDifficulty`

and an integer `d`

. The difficulty of the `ith`

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.

**Constraints:**

`1 <= jobDifficulty.length <= 300`

`0 <= jobDifficulty[i] <= 1000`

`1 <= d <= 10`

C++ Solution:

Time complexity: **O(n² * k)**

Space complexity: **O(n)**