Minimum Difficulty of a Job Schedule

Hard

Amazon

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Topics:

ArraysDynamic Programming

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.

For example:

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

As another example:

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.

How would you approach solving this problem efficiently?

Minimum Difficulty of a Job Schedule

Hard

Amazon

2 views

Topics:

ArraysDynamic Programming

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.

For example:

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

As another example:

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.

How would you approach solving this problem efficiently?

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the `jobDifficulty` array and the value of `d`? For example, what are the maximum values for `n` (length of `jobDifficulty`) and `d`?
Can the values in the `jobDifficulty` array be zero or negative?
If it's impossible to create a valid schedule (e.g., `d` is greater than the number of jobs), what value should I return?
Is `d` guaranteed to be a positive integer?
Could you define more formally how the 'difficulty of the day' is computed within each day, and how we combine these daily difficulties to determine the 'minimum difficulty of a job schedule'?

Brute Force Solution

Approach

The brute force method for the job scheduling problem involves checking every conceivable way to divide the jobs into days. We will consider every possible combination of assigning jobs to days to find the one with the least difficulty. It's like trying out every possible schedule until we find the best one.

Here's how the algorithm would work step-by-step:

First, try assigning all jobs to just one day and calculate the difficulty.
Next, try splitting the jobs into two days in every possible way, and calculate the difficulty for each of those.
Then, try splitting the jobs into three days, again considering all possible ways to divide them, and calculate the difficulty for each arrangement.
Continue this process until we've tried splitting the jobs into the maximum allowed number of days.
For each of these arrangements, calculate the total difficulty, which is derived from the hardest job of each day.
Finally, after exploring all possible ways to split the jobs into days, compare the total difficulty of each schedule and pick the one with the minimum difficulty.

Code Implementation

def minimum_difficulty_brute_force(job_difficulty, days):
    number_of_jobs = len(job_difficulty)

    # If there are fewer jobs than days, it's impossible to schedule.
    if number_of_jobs < days:
        return -1

    minimum_total_difficulty = float('inf')

    def calculate_difficulty(schedule):
        total_difficulty = 0
        for day_schedule in schedule:
            total_difficulty += max(job_difficulty[job_index] for job_index in day_schedule)
        return total_difficulty

    def generate_schedules(current_schedule, job_index_to_assign):
        nonlocal minimum_total_difficulty

        # Base case: all jobs have been assigned.
        if job_index_to_assign == number_of_jobs:
            if len(current_schedule) == days:
                difficulty = calculate_difficulty(current_schedule)
                minimum_total_difficulty = min(minimum_total_difficulty, difficulty)
            return

        # Try adding the job to the last day.
        if current_schedule:
            current_schedule[-1].append(job_index_to_assign)
            generate_schedules(current_schedule, job_index_to_assign + 1)
            current_schedule[-1].pop()

        # Try starting a new day with the job.
        # This explores splitting the jobs into more days.
        if len(current_schedule) < days:
            generate_schedules(current_schedule + [[job_index_to_assign]], job_index_to_assign + 1)

    # Start the recursion with an empty schedule.
    generate_schedules([], 0)

    # Return the minimum difficulty found.
    return minimum_total_difficulty

Big(O) Analysis

Time Complexity

O(n^d) – The algorithm explores all possible ways to divide n jobs into d days. For each day from 1 to d, the number of ways to split the jobs grows exponentially. Specifically, the number of ways to partition n jobs into d days is related to Stirling numbers of the second kind, and the computation of max difficulty for each partition adds to the operations. Thus, the complexity is approximately O(n^d) where 'n' is the number of jobs and 'd' is the number of days, due to the multiple nested loops implicitly present in exploring all partition possibilities. Since the number of ways to arrange jobs into days increases very quickly, the time complexity is exponential in the number of days.

Space Complexity

O(1) – The brute force approach, as described, primarily iterates and calculates the difficulty without storing significant intermediate results. While it explores different combinations of job schedules, it doesn't create large auxiliary data structures to hold these schedules. The algorithm mainly uses a constant number of variables to track the current split and calculate the corresponding difficulty. Therefore, the space used is independent of the number of jobs, N, resulting in constant space complexity.

Optimal Solution

Approach

The goal is to split a list of jobs into a specific number of days, minimizing the overall difficulty. We use a method that remembers the best solutions to subproblems, avoiding recomputation and leading to an efficient result.

Here's how the algorithm would work step-by-step:

Imagine you're building the schedule one day at a time, from the last day backwards to the first.
For each day, figure out which jobs could possibly be assigned to it.
The key insight: To minimize the total difficulty, each day should do the most difficult job possible among its assigned jobs.
To find the absolute best arrangement, use the following procedure: Start from the last day and compute the minimum difficulty for that day, considering all possible job assignments.
Then work backward, computing the minimum difficulty for each preceding day, remembering and reusing the minimum difficulties already calculated for the later days.
Specifically, the best schedule for any day is calculated using the best schedules for the days AFTER that day. This means we're building up from simpler subproblems to more complex ones.
Continue working backward until you find the minimum difficulty for the entire schedule, starting from the first day.

Code Implementation

def minimum_difficulty_of_a_job_schedule(job_difficulty, days_to_schedule):
    number_of_jobs = len(job_difficulty)
    if days_to_schedule > number_of_jobs:
        return -1

    # difficulty_by_job_and_day[i][d] represents the min difficulty to schedule the first i jobs in d days.
    difficulty_by_job_and_day = [[float('inf')] * (days_to_schedule + 1) for _ in range(number_of_jobs + 1)]
    difficulty_by_job_and_day[0][0] = 0

    for job_index in range(1, number_of_jobs + 1):
        for day_index in range(1, days_to_schedule + 1):
            maximum_difficulty = 0
            # Iterate backwards to find the best split
            for k_index in range(job_index, 0, -1):
                maximum_difficulty = max(maximum_difficulty, job_difficulty[k_index - 1])

                # Accumulate the maximum difficulty for each day
                difficulty_by_job_and_day[job_index][day_index] = min(difficulty_by_job_and_day[job_index][day_index], \
                                                                   difficulty_by_job_and_day[k_index - 1][day_index - 1] + maximum_difficulty)

    # Return the minimum difficulty for the entire schedule
    return difficulty_by_job_and_day[number_of_jobs][days_to_schedule]

Big(O) Analysis

Time Complexity

O(n² * d) – The algorithm utilizes dynamic programming. The outer loop iterates 'd' times, where 'd' is the number of days. Inside the outer loop, there's another loop that iterates up to 'n' times, where 'n' is the number of jobs. Within this nested loop, there is another loop that iterates a maximum of 'n' times to compute the maximum difficulty in a range of jobs assigned to a given day. Therefore, the overall time complexity is O(n * n * d), simplified to O(n² * d).

Space Complexity

O(N*D) – The solution employs dynamic programming, storing intermediate results to avoid recomputation. A table (or memoization) is maintained to store the minimum difficulty for subproblems, specifically the minimum difficulty to schedule the last 'i' jobs in 'j' days. This table has dimensions N x D, where N is the number of jobs and D is the number of days. Therefore, the auxiliary space required for this dynamic programming table is proportional to the product of N and D.

Edge Cases

Case	How to Handle
jobs array is null or empty	Return 0 if d is also 0, otherwise return -1 because no schedule is possible.
d is greater than the number of jobs	Return -1, as it's impossible to have more days than jobs.
d is 1 (single day)	The minimum difficulty is the maximum difficulty of all jobs.
jobs array contains all identical values	The algorithm should still correctly partition the jobs and calculate the minimum difficulty, potentially leading to identical difficulties on different days if 'd' allows.
jobs array contains very large difficulty values	Ensure the data type used for calculations (e.g., int) can handle the magnitude of the difficulty values to prevent integer overflow, possibly using long.
jobs array contains negative difficulty values (if not explicitly disallowed in the problem statement)	Return -1 if the job difficulties are all negative and d > 0, otherwise throw an exception or handle depending on problem constraints if negative values are disallowed; the DP array should use sufficiently large values, or initialize to infinity, not zero.
Maximum size of jobs array and/or large value of d (scalability)	Dynamic programming solution has time complexity O(nnd), so consider memory usage and runtime limits; potentially reduce space complexity with memoization.
No valid solution exists because the minimum number of jobs required per day exceeds the total number of jobs.	The algorithm returns -1 when no valid schedule can be constructed because the number of days is greater than the number of jobs.