1335. Minimum Difficulty of a Job Schedule

Problem

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

Solution

/**
 * @param {number[]} jobDifficulty
 * @param {number} d
 * @return {number}
 */
var minDifficulty = function(jobDifficulty, d) {
    var dp = Array(jobDifficulty.length).fill(0).map(() => Array(d));
    return helper(jobDifficulty, d, 0, dp);
};

var helper = function(jobs, d, index, dp) {
    if (jobs.length < d) return -1;
    if (d === 0 && index === jobs.length) return 0;
    if (d === 0) return Number.MAX_SAFE_INTEGER;
    if (dp[index][d] !== undefined) return dp[index][d];
    var min = Number.MAX_SAFE_INTEGER;
    var maxDifficulty = Number.MIN_SAFE_INTEGER;
    for (var i = index; i <= jobs.length - d; i++) {
        maxDifficulty = Math.max(maxDifficulty, jobs[i]);
        min = Math.min(min, maxDifficulty + helper(jobs, d - 1, i + 1, dp));
    }
    dp[index][d] = min;
    return min;
};

Explain:

nope.

Complexity:

• Time complexity : O(n * m).
• Space complexity : O(n * m).