1335. Minimum Difficulty of a Job Schedule
1335. Minimum Difficulty of a Job Schedule
1335. 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 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 in that day.
Given an array of integers 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.Example 4:
Input: jobDifficulty = [7,1,7,1,7,1], d = 3
Output: 15Example 5:
Input: jobDifficulty = [11,111,22,222,33,333,44,444], d = 6
Output: 843Constraints:
1 <= jobDifficulty.length <= 3000 <= jobDifficulty[i] <= 10001 <= d <= 10
思路：
题目意思是给定一个数组，每个元素代表一个任务的完成时间，再给出一个天数，让分配这些任务，每天完成的任务数不限制，耗时按照最大的计算，保证所有天数加起来耗时最小。 可以从划分的角度考虑，对于j个任务，第k天完成的任务的最小值，等于前j-k天完成的最小值，加上后面任务完成所需的最大值。那状态转移方程就可以是:
dp[i][k] = min(dp[j][k-1] + max(jobs[j+1, i])) (k - 1 <= j < i)。j个任务的取值范围有下界是因为任务数量一定要大于天数，不然题目是无解的。
代码：
golang:
func minDifficulty(jobs []int, d int) int {
m := len(jobs)
if m < d {
return -1
}
var dp = make([][]int, m + 1)
for i := range dp {
dp[i] = make([]int, d + 1)
for j := range dp[i] {
dp[i][j] = math.MaxInt16
}
}
dp[0][0] = 0
for i := 1; i <= m; i++ {
for k := 1; k <= d; k++ {
maxDay := 0
for j := i - 1; j >= k - 1; j-- {
maxDay = max(maxDay, jobs[j])
dp[i][k] = min(dp[i][k], dp[j][k-1] + maxDay)
}
}
}
return dp[m][d]
}
func max(i, j int) int {
if i > j {
return i
}
return j
}
func min(i, j int) int {
if i < j {
return i
}
return j
}