Description#
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 on that day.
You are given an integer array 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.
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:
$$
f[i][j] = \min_{k \in [1,i]} {f[k1][j1] + \max_{k \leq t \leq i} {jobDifficulty[t]}}
$$
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.
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 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]

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 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];
}
}

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 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];
}
};

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 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[k1])
f[i][j] = min(f[i][j], f[k1][j1]+mx)
}
}
}
if f[n][d] == inf {
return 1
}
return f[n][d]
}

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 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;
}
