Minimum Difficulty of a Job Schedule

Problem Description

You are given an array jobDifficulty representing the difficulty of each job and an integer d representing the number of days to schedule these jobs. You must schedule at least one job per day, and all jobs must be completed. The difficulty of a day is the maximum difficulty of the jobs done on that day. The goal is to minimize the total difficulty of the schedule, which is the sum of the difficulties of each day. If it's not possible to schedule the jobs in d days, return -1.

Naive Approach (Brute Force with Recursion)

A brute-force approach would involve trying all possible combinations of splitting the jobs into d days and calculating the total difficulty for each combination. This can be implemented using recursion.

1. Base Case: If d == 1, the difficulty is the maximum difficulty among the remaining jobs.
2. Recursive Step: Iterate through all possible split points for the first day. For each split point, calculate the maximum difficulty of jobs done on the first day and recursively calculate the minimum difficulty for the remaining days and jobs.

def minDifficulty_brute_force(jobDifficulty, d):
    n = len(jobDifficulty)
    if n < d:
        return -1

    def solve(index, days_left):
        if days_left == 1:
            return max(jobDifficulty[index:])

        min_diff = float('inf')
        curr_max = 0
        for i in range(index, n - days_left + 1):
            curr_max = max(curr_max, jobDifficulty[i])
            min_diff = min(min_diff, curr_max + solve(i + 1, days_left - 1))

        return min_diff

    return solve(0, d)

Time Complexity

The brute-force approach has exponential time complexity since it explores all possible combinations of job schedules. The complexity is roughly O(N^D), where N is the number of jobs and D is the number of days.

Space Complexity

The space complexity is O(D) due to the recursion depth.

Optimal Approach (Dynamic Programming)

To optimize the solution, we can use dynamic programming to avoid redundant calculations. We create a DP table dp[i][j], where dp[i][j] represents the minimum difficulty to schedule the first i jobs in j days.

1. Base Case: dp[i][1] is the maximum difficulty of the first i jobs.
2. Recursive Relation: dp[i][j] = min(dp[k][j-1] + max_difficulty(k+1, i)) for all k from j-1 to i-1, where max_difficulty(k+1, i) is the maximum difficulty among jobs from k+1 to i.

def minDifficulty(jobDifficulty, d):
    n = len(jobDifficulty)
    if n < d:
        return -1

    dp = [[float('inf')] * (d + 1) for _ in range(n + 1)]
    dp[0][0] = 0

    for j in range(1, d + 1):
        for i in range(j, n + 1):
            max_d = 0
            for k in range(i - 1, j - 2, -1):
                max_d = max(max_d, jobDifficulty[k])
                dp[i][j] = min(dp[i][j], dp[k][j - 1] + max_d)

    return dp[n][d]

Time Complexity

The dynamic programming approach has a time complexity of O(N^2 * D), where N is the number of jobs and D is the number of days. This is because we have three nested loops: one for days, one for jobs, and one for the split point k.

Space Complexity

The space complexity is O(N * D) due to the size of the DP table.

Edge Cases

• If the number of jobs is less than the number of days, it's impossible to schedule, so return -1.
• If the number of jobs is equal to the number of days, each job is done on a separate day, so the minimum difficulty is the sum of all job difficulties.
• If d == 1, the minimum difficulty is the maximum of all job difficulties.

Example

For jobDifficulty = [6, 5, 4, 3, 2, 1] and d = 2, the optimal solution using dynamic programming would be as follows:

The dp table would be built up, and dp[6][2] would contain the minimum difficulty, which is 7.

Summary

The optimal approach using dynamic programming provides an efficient way to solve this problem with a time complexity of O(N^2 * D) and a space complexity of O(N * D). This avoids the exponential time complexity of the brute-force approach. Handling edge cases ensures the solution is robust for all valid inputs.