LeetCode 1335. Minimum Difficulty of a Job Schedule

Question

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

Source: https://leetcode.com/problems/minimum-difficulty-of-a-job-schedule/

Solution

In this problem, jobs are dependent so we can only complete jobs sequentially. For a day, we only need to consider the start job index m and the end job index n, which means on that day we complete all jobs between job m and job n (inclusive).

cache[i][j] represents the minimum sum of difficulties when completing the first j+1 jobs in i+1 days.

In the following code block, we traverse the start job index from high to low. It gets the max difficulty of jobs on that day more efficiently.

for (int k = j; k >= i; k--) {
    dayMax = Math.max(dayMax, jobDifficulty[k]);
    cache[i][j] = Math.min(cache[i][j], cache[i - 1][k - 1] + dayMax);
}

// bottom-up 2D DP, time complexity O(nnd), space complexity O(nd)
public int minDifficulty(int[] jobDifficulty, int d) {
    // parameter validation
    if (d < 1 || jobDifficulty == null || jobDifficulty.length < 1 || jobDifficulty.length < d) {
        return -1;
    }
    int n = jobDifficulty.length; // number of jobs
    // cache for DP
    // cache[i][j] represents the min sum of difficulties when completing jobs up to j job on i day
    int[][] cache = new int[d][n];
    cache[0][0] = jobDifficulty[0];
    for (int j = 1; j < n; j++) {
        cache[0][j] = Math.max(cache[0][j - 1], jobDifficulty[j]);
    }
    for (int i = 1; i < d; i++) {
        // only use a half of the cache because we must complete at least one job per day
        for (int j = i; j < n; j++) {
            cache[i][j] = Integer.MAX_VALUE;
            int dayMax = Integer.MIN_VALUE;
            // k is the index of start job on i day
            for (int k = j; k >= i; k--) {
                dayMax = Math.max(dayMax, jobDifficulty[k]);
                cache[i][j] = Math.min(cache[i][j], cache[i - 1][k - 1] + dayMax);
            }
        }
    }
    return cache[d - 1][n - 1];
}