[Dynamic Programming] Minimum Difficulty of a Job Schedule - Medium

Dynamic Programming is a strong way to solve problems that need optimization. The Minimum Difficulty of a Job Schedule is a well-known example. The main goal of this problem is to plan a set of jobs over a number of days. We want to keep the hardest job we do on any day as low as possible. To do this, we need to find the best way to spread the jobs over the days. This will help us reach the minimum difficulty.

In this article, we will look closely at the Minimum Difficulty of a Job Schedule. We will explain the problem in detail and show how to use dynamic programming to find a good answer. We will give examples in Java, Python, and C++ to make the ideas clearer. We will also talk about how to make the dynamic programming solution better. Lastly, we will look at the complexity to see how efficient our method is. We will finish with some frequently asked questions about this topic.

  • [Dynamic Programming] Minimum Difficulty of a Job Schedule Solution Overview
  • Understanding the Problem Statement of Minimum Difficulty Job Schedule
  • Dynamic Programming Approach to Minimum Difficulty Job Schedule
  • Java Implementation of Minimum Difficulty Job Schedule
  • Python Implementation of Minimum Difficulty Job Schedule
  • C++ Implementation of Minimum Difficulty Job Schedule
  • Optimizing the Dynamic Programming Solution
  • Complexity Analysis of Minimum Difficulty Job Schedule
  • Frequently Asked Questions

If we want to learn more about dynamic programming, we can check out some related articles. You might find the Dynamic Programming: Fibonacci Number and the Dynamic Programming: Coin Change useful.

Understanding the Problem Statement of Minimum Difficulty Job Schedule

The Minimum Difficulty of a Job Schedule problem is about planning jobs over a set number of days. We want to make sure that the hardest job we do on any day is as easy as possible. We have a list of job difficulties as numbers. Our goal is to split these jobs into a certain number of days. Each day’s difficulty is the hardest job we do on that day.

Problem Constraints:

  • Input:
    • An array of integers jobDifficulty. Here, jobDifficulty[i] shows the difficulty of job i.
    • An integer d tells us how many days we have to schedule the jobs.
  • Output:
    • The smallest possible difficulty of the job schedule.

Key Points:

  • The number of days d must be less than or equal to the number of jobs we have.
  • We must schedule jobs one after the other. We cannot cut a job between days.
  • The difficulty for a day comes from the hardest job we assign to that day.

Example:

Let’s look at this example:

Input:

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

Output:

7

Explanation: - One way to schedule could be: - Day 1: Jobs [6, 5] (Difficulty = 6) - Day 2: Jobs [4, 3, 2, 1] (Difficulty = 4) - The hardest job for each day is 6 and 4. So the total is 7.

We can solve this problem well with a dynamic programming approach. We can build our answer step by step using values we have already found. This way, we can get the best solution for our job scheduling problem.

Dynamic Programming Approach to Minimum Difficulty Job Schedule

To solve the Minimum Difficulty Job Schedule problem, we can use dynamic programming. We will make a 2D table called dp. Here, dp[i][j] means the minimum difficulty of scheduling the first i jobs over j days. Our goal is to find a way to divide jobs across days while keeping the hardest job on any day as low as possible.

Steps to Implement the DP Approach:

  1. Initialization:
    • We create a dp table with size (n+1) x (d+1) where n is the number of jobs and d is the number of days.
    • We set dp[0][0] = 0 and all other entries to a large number.
  2. Filling the DP Table:
    • We loop through the days from 1 to d.
    • For each day, we loop through the jobs from 1 to n.
    • For each job, we think about splitting jobs into two parts. The last part ends with the current job.
    • We find the hardest job in the last part and update the dp table.
  3. Transition Formula:

    • For each dp[i][j], we can write the transition like this:

      dp[i][j] = min(dp[k][j-1] + maxDifficulty(k+1, i)) for all k < i

    • Here, maxDifficulty(k+1, i) finds the hardest job from k+1 to i.

  4. Base Cases:
    • If d > n, we return -1 because we can’t schedule more days than jobs.
    • If d == n, we return the total of all job difficulties since each job gets its own day.

Example Code Implementation in Python:

def minDifficulty(jobDifficulty, d):
    n = len(jobDifficulty)
    if d > n:
        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(1, n + 1):
            max_difficulty = 0
            for k in range(i - 1, -1, -1):
                max_difficulty = max(max_difficulty, jobDifficulty[k])
                dp[i][j] = min(dp[i][j], dp[k][j - 1] + max_difficulty)
    
    return dp[n][d]

# Example usage
jobDifficulty = [6, 5, 4, 3, 2, 1]
d = 2
print(minDifficulty(jobDifficulty, d))  # Output: 7

Complexity Analysis:

  • Time Complexity: O(n^2 * d) where n is the number of jobs and d is the number of days.
  • Space Complexity: O(n * d) because of the dp table.

This dynamic programming method helps us find the minimum difficulty of scheduling jobs over many days. It makes sure the hardest job on any day is as low as we can get it. If you want to learn more about similar problems in dynamic programming, you can check this Dynamic Programming - Minimum Path Sum in a Grid.

Java Implementation of Minimum Difficulty Job Schedule

To solve the Minimum Difficulty of a Job Schedule problem with Java, we use a dynamic programming approach. Our goal is to plan jobs over a set number of days. We want to keep the maximum difficulty of jobs for any single day as low as possible.

Java Code Implementation

Here is a simple Java code for the problem:

public class MinimumDifficultyJobSchedule {
    public int minDifficulty(int[] jobDifficulty, int d) {
        int n = jobDifficulty.length;
        if (n < d) return -1; // Not enough jobs for the days

        // dp[i][j] shows the minimum difficulty to schedule first i jobs in j days
        int[][] dp = new int[n + 1][d + 1];
        for (int[] row : dp) {
            Arrays.fill(row, Integer.MAX_VALUE);
        }
        dp[0][0] = 0; // Base case

        for (int j = 1; j <= d; j++) {
            for (int i = j; i <= n; i++) {
                int maxDifficulty = 0;
                for (int k = i; k >= j; k--) {
                    maxDifficulty = Math.max(maxDifficulty, jobDifficulty[k - 1]);
                    dp[i][j] = Math.min(dp[i][j], dp[k - 1][j - 1] + maxDifficulty);
                }
            }
        }

        return dp[n][d];
    }

    public static void main(String[] args) {
        MinimumDifficultyJobSchedule scheduler = new MinimumDifficultyJobSchedule();
        int[] jobDifficulty = {6, 5, 4, 3, 2, 1};
        int d = 2;
        System.out.println(scheduler.minDifficulty(jobDifficulty, d)); // Output: 7
    }
}

Explanation of the Code

  1. Input Check: The code first checks if the number of jobs is less than the number of days. If it is, we return -1 because we cannot schedule.

  2. Dynamic Programming Table Setup: We make a 2D array dp. Here, dp[i][j] shows the minimum difficulty to schedule the first i jobs over j days. We fill the table with Integer.MAX_VALUE to show that these states are not computed yet.

  3. Base Case: The base case dp[0][0] is set to 0. This means if we have no jobs scheduled, the difficulty is 0.

  4. Main Logic: We go through the days and jobs:

    • We go through the jobs backward to find the maximum job difficulty for the current day.
    • We update the dp table by checking all ways to split jobs.

  5. Result: In the end, the value in dp[n][d] gives us the minimum difficulty for scheduling all jobs over d days.

This code uses dynamic programming to find the minimum difficulty of job scheduling in a clear way. For more about dynamic programming, you can check the article on dynamic programming in job scheduling.

Python Implementation of Minimum Difficulty Job Schedule

We can solve the Minimum Difficulty of a Job Schedule problem using Python. We will use a dynamic programming method. Our goal is to schedule jobs over a certain number of days. We want to keep the maximum difficulty of jobs on any day as low as possible.

Problem Overview

We have an array of job difficulties and a number d that tells us how many days we have to schedule these jobs. The minimum difficulty works like this:

  • Each day, we can schedule one or more jobs.
  • The difficulty of a day is the hardest job scheduled that day.
  • We want to make the total difficulty as low as we can across all days.

Dynamic Programming Approach

In our dynamic programming solution, we will create a DP table. Here, dp[i][j] shows the minimum difficulty to schedule the first i jobs in j days.

Implementation

Here is the Python code that does this:

def minDifficulty(jobDifficulty, d):
    n = len(jobDifficulty)
    if n < d:
        return -1  # Not enough jobs for the days

    # Create a DP table initialized to infinity
    dp = [[float('inf')] * (d + 1) for _ in range(n + 1)]
    dp[0][0] = 0  # Base case

    for j in range(1, d + 1):  # For each day
        for i in range(j, n + 1):  # At least j jobs must be assigned for j days
            max_difficulty = 0
            for k in range(i, j - 1, -1):  # Iterate jobs from i to j
                max_difficulty = max(max_difficulty, jobDifficulty[k - 1])
                dp[i][j] = min(dp[i][j], dp[k - 1][j - 1] + max_difficulty)

    return dp[n][d]

Explanation of the Code

  • Input Handling: First, we check if we have enough jobs for the days. If not, we return -1.
  • DP Table Initialization: We create a 2D list called dp. It is filled with infinity, with size (n+1) x (d+1).
  • Base Case: We set dp[0][0] to 0. This means if we assign no jobs, the difficulty is zero.
  • Filling the DP Table:
    • We loop through each day j and each job i.
    • We find the maximum difficulty of jobs for one day using a nested loop.
    • We update the DP table with the lowest difficulty we find.

Usage

You can call the function with a list of job difficulties and the number of days:

jobs = [6, 5, 4, 3, 2, 1]
days = 2
print(minDifficulty(jobs, days))  # Output: 7

This code helps us find the minimum difficulty of scheduling jobs over the days we have using dynamic programming. If you want to learn more about dynamic programming, you can check out Dynamic Programming - Coin Change for more practice.

C++ Implementation of Minimum Difficulty Job Schedule

We will implement the Minimum Difficulty Job Schedule problem in C++. We use a dynamic programming approach. The goal is to schedule jobs over a certain number of days. We want to make the maximum difficulty of any day as low as possible. Below is the C++ code for this problem.

C++ Code

#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>

using namespace std;

class Solution {
public:
    int minDifficulty(vector<int>& jobDifficulty, int d) {
        int n = jobDifficulty.size();
        if (n < d) return -1; // Not enough jobs for the days
        
        // dp[i][j] means the minimum difficulty to schedule jobs from 0 to i in j days
        vector<vector<int>> dp(n + 1, vector<int>(d + 1, INT_MAX));
        dp[0][0] = 0; // No jobs, no difficulty
        
        for (int j = 1; j <= d; j++) {
            for (int i = j; i <= n; i++) { // We need at least j jobs in j days
                int maxDifficulty = 0;
                for (int k = i; k >= j; k--) { // Schedule jobs from k to i on day j
                    maxDifficulty = max(maxDifficulty, jobDifficulty[k - 1]);
                    dp[i][j] = min(dp[i][j], dp[k - 1][j - 1] + maxDifficulty);
                }
            }
        }
        
        return dp[n][d];
    }
};

int main() {
    Solution solution;
    vector<int> jobDifficulty = {6, 5, 4, 3, 2, 1};
    int d = 2;
    cout << "Minimum Difficulty of Job Schedule: " << solution.minDifficulty(jobDifficulty, d) << endl;
    return 0;
}

Explanation of Code

  • Input: The input has a vector jobDifficulty. This shows the difficulty of each job. There is also an integer d that shows the number of days.
  • Dynamic Programming Table: We make a 2D DP table dp. Here dp[i][j] keeps the minimum difficulty to schedule the first i jobs in j days.
  • Base Case: dp[0][0] = 0 means that if no jobs are scheduled, the difficulty is zero.
  • Transition: For each day from 1 to d, and for each job from j to n, we find the maximum difficulty of jobs for that day. We then update the DP table.
  • Output: The answer is in dp[n][d]. This shows the minimum difficulty of scheduling all jobs in d days.

This code calculates the minimum difficulty well while following the problem rules.

For more about dynamic programming, you can check articles like Dynamic Programming - Minimum Path Sum in a Grid.

<h2>Optimizing the Dynamic Programming Solution</h2>
<p>We can make the dynamic programming solution for the Minimum Difficulty of a Job Schedule problem better. We do this by using a rolling array. Instead of keeping a big 2D array, we only track the important previous states.</p>

<p>The main point is to see that to find the minimum difficulty for a day, we only need results from the day before. This lets us use a one-dimensional array to hold the minimum difficulty values. So, we reduce the space from O(n * d) to O(n). Here, n is the number of jobs and d is the number of days.</p>

<p>The better approach has these steps:</p>
<ul>
    <li>Use one array to keep the minimum difficulty up to the current day.</li>
    <li>Go through the jobs and calculate the difficulty for the current day based on the day before.</li>
    <li>Use a variable to track the hardest job for the current day.</li>
</ul>

<p>Here is a Java code for the better solution:</p>
<pre><code>public class JobScheduler {
    public int minDifficulty(int[] jobDifficulty, int d) {
        int n = jobDifficulty.length;
        if (n < d) return -1;

        int[] dp = new int[d + 1];
        for (int i = 1; i <= d; i++) {
            dp[i] = Integer.MAX_VALUE;
            for (int j = i - 1; j < n; j++) {
                int maxDifficulty = 0;
                for (int k = j; k >= i - 1; k--) {
                    maxDifficulty = Math.max(maxDifficulty, jobDifficulty[k]);
                    dp[i] = Math.min(dp[i], dp[i - 1] + maxDifficulty);
                }
            }
        }
        return dp[d];
    }
}</code></pre>

<p>In this code:</p>
<ul>
    <li>We start with a DP array. Here, <code>dp[i]</code> shows the minimum difficulty for scheduling jobs over <code>i</code> days.</li>
    <li>We go through the jobs and keep a variable for the hardest job in the current range. We update the DP array as needed.</li>
</ul>

<p>This optimization really cuts down the space we need but keeps the speed of dynamic programming methods. If you want to read more about dynamic programming, you can check out <a href="https://bestonlinetutorial.com/dynamic_programming/dynamic-programming-maximum-profit-in-job-scheduling-medium.html">Maximum Profit in Job Scheduling</a>.</p>

Complexity Analysis of Minimum Difficulty Job Schedule

The Minimum Difficulty of a Job Schedule problem is a dynamic programming problem. We need to look at time and space complexity to understand how efficient it is.

Time Complexity

We can analyze the time complexity of the Minimum Difficulty Job Schedule by looking at the dynamic programming approach. The main idea is to use a 2D DP table where:

  • n is the number of jobs
  • d is the number of days we have for scheduling

We check each job and look at possible splits for each day. This gives us the following time complexity:

  • Time Complexity: O(n^2 * d)

For each job, we check all previous jobs (up to n). For each day, we may check up to d days.

Space Complexity

The space complexity comes from the DP table that stores results. The size of this table depends on the number of jobs and days:

  • Space Complexity: O(n * d)

We need this space to keep the DP states that show the minimum difficulty for scheduling jobs over the days we have.

Example

For example, if we have 6 jobs and we need to schedule them over 3 days, the DP table will need space for 6 (jobs) * 3 (days). This gives us a space complexity of O(6 * 3) = O(18).

Conclusion

Knowing the time and space complexity helps us see if the Minimum Difficulty Job Schedule works well for larger inputs. We can think about ways to make it better. We could use a rolling array or memoization to avoid doing the same calculations again.

For more information on related dynamic programming topics, we can check these articles: Dynamic Programming - Fibonacci Number and Dynamic Programming - Minimum Path Sum.

Frequently Asked Questions

1. What is the Minimum Difficulty of a Job Schedule problem in dynamic programming?

The Minimum Difficulty of a Job Schedule problem is a well-known challenge in dynamic programming. Here, we want to schedule jobs over a few days. Our goal is to make the hardest day as easy as possible. The difficulty of a job is the most work we have to do in one day. We need to plan carefully to share the jobs well over the days.

2. How do you approach solving the Minimum Difficulty of a Job Schedule problem using dynamic programming?

To solve the Minimum Difficulty of a Job Schedule problem, we often use a recursive way with memoization or a table method. First, we need to set a state that shows the minimum difficulty for a specific job starting point and a certain day. Then, we look at all the ways to distribute jobs for that day. After that, we calculate the total difficulty for the next days.

3. What is the time complexity of the Minimum Difficulty of a Job Schedule solution?

The time complexity of the Minimum Difficulty of a Job Schedule solution depends on how many jobs and days we have. Usually, we can say the time complexity is O(n^2 * d). Here, n is the number of jobs and d is the number of days. This comes from looking at jobs and days to find the best schedule.

4. Can you implement the Minimum Difficulty of a Job Schedule in Java?

Yes, we can implement the Minimum Difficulty of a Job Schedule in Java. We need to write a dynamic programming function that keeps track of the minimum difficulty for each group of jobs and days. The code usually has a helper function that uses recursion and memoization to save results for states we have calculated before. This helps make the process faster. You can see the detailed Java code in this article for help.

5. Are there any similar dynamic programming problems I can practice to understand the Minimum Difficulty of a Job Schedule better?

Yes! To improve our understanding of dynamic programming and job scheduling, we can try similar problems. Some examples are Maximum Profit in Job Scheduling and the 0-1 Knapsack Problem. These problems have similar ideas and can help us become better at dynamic programming. For practice, we can read the article on Dynamic Programming: Maximum Profit in Job Scheduling.