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
Code :
const minDifficulty = (A, D) => {
let n = A.length, inf = Infinity, maxd;
if (n < D) return -1;
let dp = new Array(n + 1).fill(Infinity);
dp[n] = 0;
for (let d = 1; d <= D; d++) {
for (let i = 0; i <= n - d; i++) {
maxd = 0, dp[i] = inf;
for (let j = i; j <= n - d; j++) {
maxd = Math.max(maxd, A[j]);
dp[i] = Math.min(dp[i], maxd + dp[j + 1]);
}
}
}
return dp[0];
}