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Find Minimum Time to Finish All Jobs

Difficulty: Hard

Problem Description

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.

Key Insights

  • Each job must be assigned to exactly one worker.
  • The objective is to minimize the maximum working time among all workers.
  • The problem can be approached using backtracking or dynamic programming due to the constraints.
  • We can use a bitmask to represent job assignments to workers, making it easier to evaluate different combinations.

Space and Time Complexity

Time Complexity: O(k * 2^n), where n is the number of jobs and k is the number of workers. Each state can be represented in a bitmask of length n, and we evaluate k workers for each possible job assignment. Space Complexity: O(k * 2^n) for storing the state of job assignments and their corresponding maximum working time.

Solution

The problem can be solved using a backtracking approach combined with a bitmask to efficiently represent job assignments. We will maintain an array to track the current workload of each worker. For each job, we will try to assign it to each worker and recursively compute the maximum workload. The goal is to keep track of the minimum possible maximum workload encountered during these assignments.

Code Solutions

def minimumTimeRequired(jobs, k):
    n = len(jobs)
    jobs.sort(reverse=True)  # Sort jobs in descending order for better performance
    worker_times = [0] * k  # Array to track the working time of each worker
    
    def backtrack(i):
        if i == n:  # All jobs have been assigned
            return max(worker_times)  # Return the maximum workload
        
        min_time = float('inf')  # Initialize minimum time as infinity
        
        for j in range(k):
            if j > 0 and worker_times[j] == worker_times[j - 1]:
                continue  # Skip to avoid redundant calculations
            
            worker_times[j] += jobs[i]  # Assign job to worker j
            min_time = min(min_time, backtrack(i + 1))  # Recursively assign the next job
            worker_times[j] -= jobs[i]  # Backtrack to try next worker
            
        return min_time
    
    return backtrack(0)  # Start backtracking from the first job
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