Introduction

In this blog post, we will explore how to solve the LeetCode problem "1723" 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 are given an integer array jobs, where jobs[i] is the amount of time it takes to complete the ith job. There are k workers that you can assign jobs to. Each job should be assigned to exactly one worker. The working time of a worker is the sum of the time it takes to complete all jobs assigned to them. Your goal is to devise an optimal assignment such that the maximum working time of any worker is minimized. Return the minimum possible maximum working time of any assignment. Example 1: Input: jobs = [3,2,3], k = 3 Output: 3 Explanation: By assigning each person one job, the maximum time is 3. Example 2: Input: jobs = [1,2,4,7,8], k = 2 Output: 11 Explanation: Assign the jobs the following way: Worker 1: 1, 2, 8 (working time = 1 + 2 + 8 = 11) Worker 2: 4, 7 (working time = 4 + 7 = 11) The maximum working time is 11. Constraints: 1 <= k <= jobs.length <= 12 1 <= jobs[i] <= 107

Explanation

Here's a breakdown of the approach and the Python code:

• Binary Search: We perform a binary search on the possible range of maximum working times. The lower bound is the maximum job duration, and the upper bound is the sum of all job durations.

• Backtracking: For each 'mid' value (potential maximum working time) in the binary search, we use a backtracking algorithm to check if it's possible to assign jobs to workers such that no worker exceeds the 'mid' working time.

• Optimization: The backtracking is optimized by pruning branches early. If a worker's current load exceeds 'mid', we stop exploring that branch. We also avoid assigning the same job to a worker multiple times in the same branch.

• Runtime Complexity: O(2ⁿ n log(sum(jobs))), where n is the number of jobs. The 2ⁿ comes from the backtracking, n is from copying arrays to prevent mutation during recursive calls and log(sum(jobs)) comes from the binary search. Storage Complexity: O(n + k), where n is the recursion depth (number of jobs) and k is the number of workers.

Code

    def minimumTimeRequired(jobs, k):
    """
    Finds the minimum possible maximum working time of any assignment of jobs to k workers.

    Args:
        jobs: A list of integers representing the time it takes to complete each job.
        k: The number of workers.

    Returns:
        The minimum possible maximum working time of any assignment.
    """

    def is_possible(max_time):
        """
        Checks if it's possible to assign jobs to workers such that no worker exceeds max_time.
        """
        workers = [0] * k

        def backtrack(job_index):
            if job_index == len(jobs):
                return True

            for i in range(k):
                if workers[i] + jobs[job_index] <= max_time:
                    workers[i] += jobs[job_index]
                    if backtrack(job_index + 1):
                        return True
                    workers[i] -= jobs[job_index]
                    # optimization: if the worker is currently empty, we can skip further assignments since
                    # a worker is already idle and it didn't work
                    if workers[i] == 0:
                        break  # critical optimization!

            return False

        return backtrack(0)

    left, right = max(jobs), sum(jobs)
    ans = right

    while left <= right:
        mid = (left + right) // 2
        if is_possible(mid):
            ans = mid
            right = mid - 1
        else:
            left = mid + 1

    return ans