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

The optimal solution has a time complexity of O(n*n*d).

What data structures are used to solve Minimum Difficulty of a Job Schedule?

This problem uses Array, Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Difficulty of a Job Schedule?

Insight 1: We need to use dynamic programming because we are looking for the minimum cost (difficulty) among many possible schedules. We can break the problem into smaller subproblems. Insight 2: The problem has overlapping subproblems. For instance, the optimal solution for the last 'k' days might be used in calculating the solution for the last 'k+1' days.

Which companies ask LeetCode 1335. Minimum Difficulty of a Job Schedule in interviews?

This problem is frequently asked at Turvo.

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Minimum Difficulty of a Job Schedule - LeetCode 1335 Solution

Q: How to solve Minimum Difficulty of a Job Schedule?

The solution uses a top-down dynamic programming approach with memoization. A recursive function `dp(i, k)` is defined, which represents the minimum difficulty of scheduling the jobs from index 'i' to the end of the `jobDifficulty` array over 'k' days. The function calculates the difficulty of the current day by iterating through possible jobs for the current day and recursively calls `dp` for the remaining jobs and remaining days. The results are memoized using `@functools.lru_cache` in Python and a 2D array `memo` in Java and C++ (and a similar manual implementation in C) to avoid recomputation of overlapping subproblems.

Q: What is the space complexity of Minimum Difficulty of a Job Schedule?

The space complexity is O(n*d).

Minimum Difficulty of a Job Schedule - Complete Solution Guide

Minimum Difficulty of a Job Schedule is LeetCode problem 1335, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You want to schedule a list of jobs in d days. Jobs are dependent (i.e To work on the i th 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 i th job is jobDifficulty[i] . Return the minimum di

Detailed Explanation

The problem asks us to find the minimum difficulty of scheduling a list of jobs over 'd' days. We must complete at least one job per day, and the difficulty of a day is the maximum difficulty of any job done on that day. The total difficulty is the sum of the difficulties of each day. The jobs must be done in order, meaning job 'i' can only be done after jobs 0 to i-1 are finished. If it's impossible to schedule all jobs in 'd' days (e.g., more days than jobs), we should return -1.

Solution Approach

The solution uses a top-down dynamic programming approach with memoization. A recursive function `dp(i, k)` is defined, which represents the minimum difficulty of scheduling the jobs from index 'i' to the end of the `jobDifficulty` array over 'k' days. The function calculates the difficulty of the current day by iterating through possible jobs for the current day and recursively calls `dp` for the remaining jobs and remaining days. The results are memoized using `@functools.lru_cache` in Python and a 2D array `memo` in Java and C++ (and a similar manual implementation in C) to avoid recomputation of overlapping subproblems.

Step-by-Step Algorithm

Step 1: Check if the number of jobs 'n' is less than the number of days 'd'. If so, return -1, as it's impossible to schedule at least one job per day.
Step 2: Define a recursive function `dp(i, k)` to find the minimum difficulty of scheduling jobs from index 'i' using 'k' days.
Step 3: Base case: If k == 1, it means we have only one day left. The difficulty is simply the maximum job difficulty from index 'i' to the end of the array.
Step 4: For each day, iterate through possible jobs to do on that day (from index 'i' to 'n - k').
Step 5: Calculate the difficulty of that day (maximum difficulty of jobs done on that day).
Step 6: Recursively call `dp(j + 1, k - 1)` to find the minimum difficulty for the remaining jobs and remaining days.
Step 7: Add the current day's difficulty to the result of the recursive call, and update the minimum difficulty found so far.
Step 8: Memoize the result of `dp(i, k)` to avoid recomputation.
Step 9: The function call dp(0, d) will then give the solution to the original problem.

Key Insights

Insight 1: We need to use dynamic programming because we are looking for the minimum cost (difficulty) among many possible schedules. We can break the problem into smaller subproblems.
Insight 2: The problem has overlapping subproblems. For instance, the optimal solution for the last 'k' days might be used in calculating the solution for the last 'k+1' days.
Insight 3: Memoization (or tabulation) is essential to avoid recomputing the same subproblems, making the solution efficient.

Complexity Analysis

Time Complexity: O(n*n*d)

Space Complexity: O(n*d)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Turvo.

Minimum Difficulty of a Job Schedule | LeetCode 1335

Hard

Array Dynamic Programming

Time: O(n*n*d)Space: O(n*d)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    private int[] jobDifficulty;
    private int d;
    private Integer[][] memo;
    private int n;

    public int minDifficulty(int[] jobDifficulty, int d) {
        this.jobDifficulty = jobDifficulty;
        this.d = d;
        this.n = jobDifficulty.length;
        
        if (n < d) {
            return -1;
        }
        
        memo = new Integer[n][d + 1];
        return dp(0, d);
    }

    private int dp(int i, int k) {
        if (memo[i][k] != null) {
            return memo[i][k];
        }
        
        if (k == 1) {
            int maxDifficulty = 0;
            for (int j = i; j < n; j++) {
                maxDifficulty = Math.max(maxDifficulty, jobDifficulty[j]);
            }
            memo[i][k] = maxDifficulty;
            return maxDifficulty;
        }

        int res = Integer.MAX_VALUE;
        int max_d_today = 0;
        for (int j = i; j <= n - k; j++) {
            max_d_today = Math.max(max_d_today, jobDifficulty[j]);
            res = Math.min(res, max_d_today + dp(j + 1, k - 1));
        }

        memo[i][k] = res;
        return res;
    }
}

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Problem Info

Description

Problem 1335: Minimum Difficulty of a Job Schedule - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

Turvo

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1335

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Check if the number of jobs 'n' is less than the number of days 'd'. If so, return -1, as it's impossible to schedule at least one job per day.
2
Step 2: Define a recursive function `dp(i, k)` to find the minimum difficulty of scheduling jobs from index 'i' using 'k' days.
3
Step 3: Base case: If k == 1, it means we have only one day left. The difficulty is simply the maximum job difficulty from index 'i' to the end of the array.
4
Step 4: For each day, iterate through possible jobs to do on that day (from index 'i' to 'n - k').
5
Step 5: Calculate the difficulty of that day (maximum difficulty of jobs done on that day).
6
Step 6: Recursively call `dp(j + 1, k - 1)` to find the minimum difficulty for the remaining jobs and remaining days.
7
Step 7: Add the current day's difficulty to the result of the recursive call, and update the minimum difficulty found so far.
8
Step 8: Memoize the result of `dp(i, k)` to avoid recomputation.
9
Step 9: The function call dp(0, d) will then give the solution to the original problem.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*n*d)

💾

Space Complexity

O(n*d)

📝

Why?

The time complexity is O(n^2 * d) because the dp function is called n*d times and for each call it iterates through n possible values for j. The space complexity is O(n*d) due to the memoization table that stores the solutions to the subproblems to prevent recomputation. The reason C also uses (n+1)*(d+1) space is to account for cases where 'i' can be n and 'k' can be d, for proper indexing without causing out-of-bounds errors.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Bottom-up Dynamic Programming (Tabulation)

Instead of recursion with memoization, we can use tabulation to build the DP table iteratively. We start with the base case (d=1) and then build up the table for d=2, d=3, and so on.

Time: O(n*n*d)Space: O(n*d)

Trade-offs: Bottom-up DP can sometimes be slightly faster because it avoids the overhead of recursive function calls. However, it might be less intuitive to implement than top-down DP. In many cases, the performance difference is negligible.

✨

Key Insights

1Insight 1: We need to use dynamic programming because we are looking for the minimum cost (difficulty) among many possible schedules. We can break the problem into smaller subproblems.
2Insight 2: The problem has overlapping subproblems. For instance, the optimal solution for the last 'k' days might be used in calculating the solution for the last 'k+1' days.
3Insight 3: Memoization (or tabulation) is essential to avoid recomputing the same subproblems, making the solution efficient.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `jobDifficulty.length < d`. The solution correctly handles this edge case by returning -1 because it's impossible to schedule at least one job per day.
Edge case 2: `d = 1`. In this case, the minimum difficulty is the maximum difficulty among all jobs. The base case `k == 1` in the `dp` function correctly handles this.
Edge case 3: Large job difficulties leading to integer overflow. In C, proper bounds checking `next_val < INT_MAX - max_d_today` is performed to avoid overflow.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where `jobDifficulty.length < d`, which leads to incorrect results or runtime errors.

✗Common mistake 2: Not using memoization, which results in exponential time complexity and a time limit exceeded error.

✗Common mistake 3: Incorrect base case definition. The base case needs to correctly calculate the difficulty when only one day is remaining.

✗Common mistake 4: Off-by-one errors in indexing, especially when iterating through the jobs for a given day or calculating the remaining jobs for the recursive call.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Task scheduling: This algorithm can be adapted to schedule tasks with varying difficulties over a limited number of days, minimizing the overall 'difficulty' or cost of the schedule.

🏦Real-world scenario 2: Resource allocation: Similar to task scheduling, this algorithm can be used to allocate resources to different projects or tasks over a period of time, minimizing the total cost or maximizing the total efficiency.

🎮Real-world scenario 3: Project management: Breaking down a project into smaller tasks and scheduling them over a timeline to minimize the overall project difficulty or risk.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 15, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Difficulty of a Job Schedule"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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