1335. Minimum Difficulty of a Job Schedule

Hard
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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:
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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

解題

func minDifficulty(jobDifficulty []int, d int) int {
n:= len(jobDifficulty)
if (d > n) { return -1 }
var dp = make([][]int, d+1)
for i := range dp {
dp[i] = make([]int, n+1)
for j:= 0; j<n+1; j++ {
dp[i][j] = math.MaxInt
}
}
dp[0][0] = 0
for day:= 1; day<=d; day++ {
for task:= day; task<=n; task++ {
localMax := jobDifficulty[task-1];
for s:=task; s>=day; s-- {
if jobDifficulty[s-1]>localMax {
localMax = jobDifficulty[s-1]
}
if (dp[day-1][s-1] != math.MaxInt ) {
if (dp[day-1][s-1]+localMax)<dp[day][task]{
dp[day][task] = dp[day-1][s-1]+localMax
}
}
}
}
}
return dp[d][n];
}go