How do you solve Minimum Difficulty of a Job Schedule on LeetCode?

Minimum Difficulty of a Job Schedule (LeetCode #1335) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n² * d) time complexity and O(n * d) space complexity. Key insight: Understanding how to partition the jobs effectively is crucial for minimizing difficulty.

What is the brute force solution for Minimum Difficulty of a Job Schedule?

The brute force approach for Minimum Difficulty of a Job Schedule is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying all possible ways to split the jobs into `d` days. For each split, we calculate the maximum difficulty of jobs for each day and sum these maximums to find the total difficulty. The optimal Optimal Solution approach improves this to O(n² * d).

What algorithm or data structure is used to solve Minimum Difficulty of a Job Schedule?

Minimum Difficulty of a Job Schedule uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Minimum Difficulty of a Job Schedule?

Always clarify the constraints and edge cases before diving into the solution. Explain your thought process clearly as you work through the problem, especially when transitioning from brute force to optimal solutions. Practice dynamic programming problems to get comfortable with state representation and transitions.

What are common mistakes when solving Minimum Difficulty of a Job Schedule?

Failing to check if the number of jobs is less than the number of days, which leads to incorrect assumptions. Not considering all possible partitions, which can lead to suboptimal solutions.

#1335

Minimum Difficulty of a Job Schedule

Hard

ArrayDynamic ProgrammingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n² * d)
Space	O(1)	O(n * d)

💡

Intuition

Time O(n² * d)Space O(n * d)

Using dynamic programming, we can break down the problem into smaller subproblems. We maintain a DP table where each entry represents the minimum difficulty of scheduling jobs up to a certain index with a given number of days left.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP table where dp[i][j] represents the minimum difficulty of scheduling the first i jobs in j days.
2Step 2: Iterate through the number of days and for each day, calculate the maximum job difficulty for the jobs scheduled on that day.
3Step 3: Update the DP table based on the previous day's minimum difficulty and the maximum job difficulty for the current day.

solution.py16 lines

1# Full working Python code
2import sys
3
4def minDifficulty(jobDifficulty, d):
5    n = len(jobDifficulty)
6    if n < d:
7        return -1
8    dp = [[sys.maxsize] * (d + 1) for _ in range(n + 1)]
9    dp[0][0] = 0
10    for j in range(1, d + 1):
11        for i in range(j, n + 1):
12            max_difficulty = 0
13            for k in range(i, j - 1, -1):
14                max_difficulty = max(max_difficulty, jobDifficulty[k - 1])
15                dp[i][j] = min(dp[i][j], dp[k - 1][j - 1] + max_difficulty)
16    return dp[n][d]

ℹ

Complexity note: The time complexity is O(n² * d) because for each day, we may need to check all jobs up to that day, and we do this for each day, leading to a quadratic relationship with respect to the number of jobs.

1Understanding how to partition the jobs effectively is crucial for minimizing difficulty.
2Dynamic programming allows us to build solutions incrementally, making it easier to manage complex dependencies.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.