Minimum Difficulty of a Job Schedule

class Solution {
    public int findMinDifficulty(int[] jobs, int days) {
        int jobCount = jobs.length;
        if (jobCount < days) {
            return -1;
        }

        int[] prevMinDiff = new int[jobCount];
        int[] currMinDiff = new int[jobCount];
        int[] swap;
        
        Arrays.fill(prevMinDiff, 1000);
        Deque<Integer> indices = new ArrayDeque<>();

        for (int dayCounter = 0; dayCounter < days; ++dayCounter) {
            indices.clear();
            for (int j = dayCounter; j < jobCount; j++) {
                currMinDiff[j] = j > 0 ? prevMinDiff[j - 1] + jobs[j] : jobs[j];
                while (!indices.isEmpty() && jobs[indices.peek()] <= jobs[j]) {
                    int index = indices.pop();
                    int difficultyIncrease = jobs[j] - jobs[index];
                    currMinDiff[j] = Math.min(currMinDiff[j], currMinDiff[index] + difficultyIncrease);
                }

                if (!indices.isEmpty()) {
                    currMinDiff[j] = Math.min(currMinDiff[j], currMinDiff[indices.peek()]);
                }

                indices.push(j);
            }
            swap = prevMinDiff;
            prevMinDiff = currMinDiff;
            currMinDiff = swap;
        }

        return prevMinDiff[jobCount - 1];
    }
}

The Java code provided addresses the problem of finding the minimum difficulty of a job schedule for a given number of days. The solution involves dynamic programming to efficiently manage and compute the minimum possible difficulty over specified days.

The key strategy in the code is as follows:

• First, a check ensures there are enough jobs to fill each day; if not, the function returns -1.
• Two arrays, prevMinDiff and currMinDiff, are introduced to hold temporal results for the minimum difficulty computations.
• The function uses a sliding window approach, maintained by a deque (indices), to keep track of indices in the job list that offer the least increase in difficulty if selected for the current day's end.
• For each day in the job scheduling (outer loop over days), the code adjusts the deque and updates the challenges in the currMinDiff array based on the minimum difficulty encountered using previously computed results from prevMinDiff.
• After completing the inner loop over job indices, prevMinDiff and currMinDiff arrays are swapped, storing the current day's results to be used as "previous" for the next day calculation.

By the end of the designated number of days, the element prevMinDiff[jobCount - 1] will reveal the minimum difficulty of the entire schedule. This efficient approach ensures that each day's computation considers the best arrangement up to that point, optimizing both time and space complexity using dynamic programming techniques.

class Solution:
    def calculateMinDifficulty(self, jobDiff, days):
        total_jobs = len(jobDiff)
        if total_jobs < days:
            return -1

        previous_day_diff, current_day_diff = [float('inf')] * total_jobs, [float('inf')] * total_jobs

        for day_idx in range(days):
            monotonous_stack = []
            for job_idx in range(day_idx, total_jobs):

                if job_idx == 0:
                    current_day_diff[job_idx] = jobDiff[0]
                else:
                    current_day_diff[job_idx] = previous_day_diff[job_idx - 1] + jobDiff[job_idx]

                while monotonous_stack and jobDiff[monotonous_stack[-1]] <= jobDiff[job_idx]:

                    popped_idx = monotonous_stack.pop()
                    difficulty_increase = jobDiff[job_idx] - jobDiff[popped_idx]
                    current_day_diff[job_idx] = min(current_day_diff[job_idx], current_day_diff[popped_idx] + difficulty_increase)

                if monotonous_stack:
                    current_day_diff[job_idx] = min(current_day_diff[job_idx], current_day_diff[monotonous_stack[-1]])

                monotonous_stack.append(job_idx)

            previous_day_diff, current_day_diff = current_day_diff, previous_day_diff

        return previous_day_diff[-1]

The given Python solution strategically determines the minimum difficulty of scheduling job tasks over a specified number of days. The function calculateMinDifficulty takes two parameters: jobDiff, an array where each element represents the difficulty of a job, and days, the number of days to schedule these jobs.

• Initialize the length of jobDiff to total_jobs.
• If the number of jobs is less than the required days, return -1 since it's not possible to schedule them.
• Use dynamic programming with two lists, previous_day_diff and current_day_diff, initialized with float('inf'), to keep track of the minimum difficulty at each step.
• Loop through each day using day_idx and through each job using job_idx from day_idx to total_jobs.
• Use a monotonous stack to efficiently calculate the minimum difficulty:
  • For the start of a segment, set the difficulty directly from jobDiff.
  • For other jobs, compute the minimum difficulty by comparing current difficulty with the potential difficulty if this job was included in the current day’s task.
  • While jobs in the stack have difficulty less than the current job, update the current day's difficulty considering the increased difficulty due to the harder current job.
  • The stack helps in maintaining jobs in descending order of their difficulty, ensuring that every job is only computed with the more difficult ones preceding it.
• After scanning through all jobs for the day, swap previous_day_diff and current_day_diff to proceed to the next day.
• After processing all days, the last element in previous_day_diff gives the minimum difficulty for scheduling all jobs within the specified days.