Introduction

In this blog post, we will explore how to solve the LeetCode problem "1335" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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. Example 1: Input: 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 Example 2: Input: 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. Example 3: Input: jobDifficulty = [1,1,1], d = 3 Output: 3 Explanation: The schedule is one job per day. total difficulty will be 3. Constraints: 1 <= jobDifficulty.length <= 300 0 <= jobDifficulty[i] <= 1000 1 <= d <= 10

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: Utilize dynamic programming to explore all possible job schedules, building up solutions from smaller subproblems. dp[i][j] stores the minimum difficulty to schedule the first i jobs in j days.
• State Transition: For each dp[i][j], iterate through all possible split points k (where the last day's work starts). The transition calculates the maximum difficulty of the last day (jobs k+1 to i) and adds it to the minimum difficulty of scheduling the first k jobs in j-1 days.

• Base Cases and Constraints: Handle the base cases for single-day schedules and the constraint that at least one job must be done each day. Return -1 if a schedule is impossible.

• Runtime Complexity: O(n*n*d)

• Storage Complexity: O(n*d)

Code

    def minDifficulty(jobDifficulty, d):
    n = len(jobDifficulty)

    if n < d:
        return -1

    dp = [[float('inf')] * (d + 1) for _ in range(n + 1)]
    dp[0][0] = 0

    for i in range(1, n + 1):
        for j in range(1, min(i, d) + 1):
            max_difficulty = 0
            for k in range(i - 1, j - 2, -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]