# 1335. Minimum Difficulty of a Job Schedule

## Description

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

## Solutions

Solution 1: Dynamic Programming

We define $f[i][j]$ as the minimum difficulty to finish the first $i$ jobs within $j$ days. Initially $f[0][0] = 0$, and all other $f[i][j]$ are $\infty$.

For the $j$-th day, we can choose to finish jobs $[k,..i]$ on this day. Therefore, we have the following state transition equation:

$f[i][j] = \min_{k \in [1,i]} \{f[k-1][j-1] + \max_{k \leq t \leq i} \{jobDifficulty[t]\}\}$

The final answer is $f[n][d]$.

The time complexity is $O(n^2 \times d)$, and the space complexity is $O(n \times d)$. Here $n$ and $d$ are the number of jobs and the number of days respectively.

• class Solution {
public int minDifficulty(int[] jobDifficulty, int d) {
final int inf = 1 << 30;
int n = jobDifficulty.length;
int[][] f = new int[n + 1][d + 1];
for (var g : f) {
Arrays.fill(g, inf);
}
f[0][0] = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= Math.min(d, i); ++j) {
int mx = 0;
for (int k = i; k > 0; --k) {
mx = Math.max(mx, jobDifficulty[k - 1]);
f[i][j] = Math.min(f[i][j], f[k - 1][j - 1] + mx);
}
}
}
return f[n][d] >= inf ? -1 : f[n][d];
}
}

• class Solution {
public:
int minDifficulty(vector<int>& jobDifficulty, int d) {
int n = jobDifficulty.size();
int f[n + 1][d + 1];
memset(f, 0x3f, sizeof(f));
f[0][0] = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= min(d, i); ++j) {
int mx = 0;
for (int k = i; k; --k) {
mx = max(mx, jobDifficulty[k - 1]);
f[i][j] = min(f[i][j], f[k - 1][j - 1] + mx);
}
}
}
return f[n][d] == 0x3f3f3f3f ? -1 : f[n][d];
}
};

• class Solution:
def minDifficulty(self, jobDifficulty: List[int], d: int) -> int:
n = len(jobDifficulty)
f = [[inf] * (d + 1) for _ in range(n + 1)]
f[0][0] = 0
for i in range(1, n + 1):
for j in range(1, min(d + 1, i + 1)):
mx = 0
for k in range(i, 0, -1):
mx = max(mx, jobDifficulty[k - 1])
f[i][j] = min(f[i][j], f[k - 1][j - 1] + mx)
return -1 if f[n][d] >= inf else f[n][d]


• func minDifficulty(jobDifficulty []int, d int) int {
n := len(jobDifficulty)
f := make([][]int, n+1)
const inf = 1 << 30
for i := range f {
f[i] = make([]int, d+1)
for j := range f[i] {
f[i][j] = inf
}
}
f[0][0] = 0
for i := 1; i <= n; i++ {
for j := 1; j <= min(d, i); j++ {
mx := 0
for k := i; k > 0; k-- {
mx = max(mx, jobDifficulty[k-1])
f[i][j] = min(f[i][j], f[k-1][j-1]+mx)
}
}
}
if f[n][d] == inf {
return -1
}
return f[n][d]
}

• function minDifficulty(jobDifficulty: number[], d: number): number {
const n = jobDifficulty.length;
const inf = 1 << 30;
const f: number[][] = new Array(n + 1).fill(0).map(() => new Array(d + 1).fill(inf));
f[0][0] = 0;
for (let i = 1; i <= n; ++i) {
for (let j = 1; j <= Math.min(d, i); ++j) {
let mx = 0;
for (let k = i; k > 0; --k) {
mx = Math.max(mx, jobDifficulty[k - 1]);
f[i][j] = Math.min(f[i][j], f[k - 1][j - 1] + mx);
}
}
}
return f[n][d] < inf ? f[n][d] : -1;
}