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: 15

Example 5:

Input: jobDifficulty = [11,111,22,222,33,333,44,444], d = 6
Output: 843

 

Constraints:


Solution:

class Solution {
    Integer[][] dp;
    
    public int minDifficulty(int[] jobDifficulty, int d) {
        int n = jobDifficulty.length;
        if (d > n) return -1;
        int min = 1000 * 300;
        int maxDifficulty = 0;
        dp = new Integer[n + 1][d + 1];
        if (d == 1) {
            for (int i = 0; i < jobDifficulty.length; i ++) {
                maxDifficulty = Math.max(maxDifficulty, jobDifficulty[i]);
            }
            return maxDifficulty;
        }
        for (int i = 0; i < n - d + 1; i ++) {
            maxDifficulty = Math.max(maxDifficulty, jobDifficulty[i]);
            // System.out.println(maxDifficulty + "," + helper(jobDifficulty, i + 1, d - 1));
            min = Math.min(min, maxDifficulty + helper(jobDifficulty, i + 1, d - 1));
        }
        return min;
    }
    
    private int helper(int[] jobs, int start, int d) {
        if (dp[start][d] != null) return dp[start][d];
        if (start == jobs.length) return 0;
        if (d == 1) {
            int maxDifficulty = 0;
            for (int i = start; i < jobs.length; i ++) {
                maxDifficulty = Math.max(maxDifficulty, jobs[i]);
            }
            return dp[start][d] = maxDifficulty;
        }
        int min = 1000 * 300;
        int maxDifficulty = 0;
        for (int i = start; i < jobs.length - d + 1; i ++) {
            maxDifficulty = Math.max(maxDifficulty, jobs[i]);
            min = Math.min(min, maxDifficulty + helper(jobs, i + 1, d - 1));
        }        
        return dp[start][d] = min;
    }
}