1335. Minimum Difficulty of a Job Schedule

Hard

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

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