README.md

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 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 (Rust)

1. Solution

implSolution{pubfnmin_difficulty(job_difficulty:Vec<i32>,d:i32) -> i32{let n = job_difficulty.len();let d = d asusize;if n < d {return -1;}letmut dp = vec![vec![i32::MAX; n + 1]; d + 1];letmut max_difficulty = 0;for i in0..=n - d { max_difficulty = max_difficulty.max(job_difficulty[i]); dp[1][i + 1] = max_difficulty;}for i in1..d {for j in i..=n + i - d { max_difficulty = 0;for k in1..=n + i + 1 - j - d { max_difficulty = max_difficulty.max(job_difficulty[j + k - 1]); dp[i + 1][j + k] = dp[i + 1][j + k].min(dp[i][j] + max_difficulty);}}} dp[d][n]}}