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:
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;
}
}