#1335

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Minimum Difficulty of a Job Schedule

Schedule jobs into multiple days to minimize the difficulty of the schedule using dynamic programming and state transitions.

array dynamic programming

Problem Statement

Given an array of integers representing job difficulties, your task is to schedule these jobs into exactly d days. The constraint is that the jobs must be completed in order, and every day must have at least one job. The difficulty of a schedule is the sum of the maximum difficulty of jobs done each day.

If it's impossible to split the jobs into d days (i.e., you can't complete at least one job per day), return -1. Otherwise, return the minimum possible difficulty of the schedule. The challenge can be efficiently approached using dynamic programming and trying various splits of the job array.

Examples

Example 1

Input: jobDifficulty = [6,5,4,3,2,1], d = 2

Output: 7

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

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

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 Approach

State Transition Dynamic Programming

The solution is based on dynamic programming (DP), where the state represents the minimum difficulty for completing the jobs up to a certain index while using a specific number of days. We recursively calculate the possible splits and track the minimum difficulty by updating our DP table.

Iterate Over All Possible Cuts

For each possible cut of the job array, we divide the jobs into groups, and for each group, we calculate the maximum difficulty. This allows us to minimize the total difficulty by trying all possible combinations of splits.

Handle Base Cases and Constraints

Base cases include scenarios where d is greater than the number of jobs, resulting in no valid solution. Also, the DP approach efficiently handles these edge cases by checking the job splits early on, avoiding unnecessary calculations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexities depend on the final approach, which primarily involves calculating all possible splits and storing them in a DP table. Typically, the time complexity is O(n * d^2) and the space complexity is O(n * d), where n is the length of the job array and d is the number of days.

What Interviewers Usually Probe

Tests for edge cases where d is greater than the number of jobs.
Challenges in efficiently computing the minimum difficulty when dividing the jobs into groups.
The ability to understand and implement dynamic programming with state transitions in job scheduling.

Common Pitfalls or Variants

Common pitfalls

Overcomplicating the solution by not efficiently handling the DP state transitions.
Forgetting to handle the base case where d is greater than the number of jobs.
Not updating the DP table correctly while iterating through possible splits, leading to incorrect difficulty calculations.

Follow-up variants

Varying the size of the job array, potentially increasing complexity with larger inputs.
Changing the number of days d, which requires adjusting the approach for splitting the job array.
Allowing different job difficulties to test how the algorithm handles varied input distributions.

FAQ

How do I minimize the job schedule difficulty?

You can minimize the job schedule difficulty by dividing the jobs into d days, using dynamic programming to compute the optimal splits and minimize the difficulty of each schedule.

Why does dynamic programming work for this problem?

Dynamic programming works because we can break down the problem into subproblems of calculating the minimum difficulty for each split, building the solution progressively.

What happens if `d` is greater than the number of jobs?

If d is greater than the number of jobs, it is impossible to schedule all jobs, as each day must have at least one job, and the function will return -1.

How does state transition help with solving this problem?

State transition allows us to track the minimal difficulty across different splits of the job array, ensuring that we calculate the least possible difficulty as we divide the jobs into d days.

What is the time complexity of this problem?

The time complexity is typically O(n * d^2), where n is the number of jobs and d is the number of days, due to the DP approach iterating through all possible splits.

terminal

Solution

Solution 1: Dynamic Programming

We define $f[i][j]$ as the minimum difficulty to finish the first $i$ jobs within $j$ days. Initially $f[0][0] = 0$, and all other $f[i][j]$ are $\infty$.

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class Solution:
    def minDifficulty(self, jobDifficulty: List[int], d: int) -> int:
        n = len(jobDifficulty)
        f = [[inf] * (d + 1) for _ in range(n + 1)]
        f[0][0] = 0
        for i in range(1, n + 1):
            for j in range(1, min(d + 1, i + 1)):
                mx = 0
                for k in range(i, 0, -1):
                    mx = max(mx, jobDifficulty[k - 1])
                    f[i][j] = min(f[i][j], f[k - 1][j - 1] + mx)
        return -1 if f[n][d] >= inf else f[n][d]

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