1335. Minimum Difficulty of a Job Schedule

Description

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

Solution: Dynamic Programming

• Time complexity: O(n²d)
• Space complexity: O(nd)

 

JavaScript

/**
 * @param {number[]} jobDifficulty
 * @param {number} d
 * @return {number}
 */
const minDifficulty = function (jobDifficulty, d) {
  const n = jobDifficulty.length;

  if (n < d) return -1;
  const memo = Array.from({ length: n }, () => new Array(d).fill(-1));
  const maxDifficultyJobs = Array.from({ length: n }, () => 0);
  let currentMaxDifficulty = 0;

  for (let index = n - 1; index >= 0; index--) {
    currentMaxDifficulty = Math.max(jobDifficulty[index], currentMaxDifficulty);
    maxDifficultyJobs[index] = currentMaxDifficulty;
  }

  const finishJobs = (job, day) => {
    if (day === d) return maxDifficultyJobs[job];
    if (memo[job][day] !== -1) return memo[job][day];
    let maxDifficulty = 0;
    let result = Number.MAX_SAFE_INTEGER;

    for (let index = job; index < n - (d - day); index++) {
      maxDifficulty = Math.max(jobDifficulty[index], maxDifficulty);

      const difficulty = maxDifficulty + finishJobs(index + 1, day + 1);

      result = Math.min(difficulty, result);
    }
    memo[job][day] = result;

    return result;
  };

  return finishJobs(0, 1);
};