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:

• 1 <= jobDifficulty.length <= 300
• 0 <= jobDifficulty[i] <= 1000
• 1 <= d <= 10

Solution: DP

dp[i][k] := min difficulties to schedule jobs 1~i in k days.

Schedule 1 ~ j in k – 1 days and schedule j + 1 ~ i in 1 day.

Init: dp[0][0] = 0
Transition: dp[i][k] := min(dp[j][k -1] + max(jobs[j + 1 ~ i]), k – 1 <= j < i
Answer: dp[n][d]

Time complexity: O(n^2*d)
Space complexity: O(n)

// Author: Huahua
// Author: Huahua
class Solution {
public:
  int minDifficulty(vector<int>& jobs, int d) {
    const int n = jobs.size();
    if (d > n) return -1;    
    vector<vector<int>> dp(n + 1, vector<int>(d + 1, INT_MAX / 2));
    
    dp[0][0] = 0;
    for (int i = 1; i <= n; ++i)      
      for (int j = i - 1, md = 0; j >= 0; --j) {
        md = max(md, jobs[j]);
        for (int k = 1; k <= min(i, d); ++k)                    
          dp[i][k] = min(dp[i][k], dp[j][k - 1] + md);
      }
    
    return dp[n][d];
  }
};

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