1723. Find Minimum Time to Finish All Jobs

Problem Description

In this problem, we are given an array called jobs, where each entry jobs[i] represents the time it takes to complete the i-th job. Additionally, there are k workers available to assign these jobs to. Each job must be assigned to one and only one worker. The "working time" for a worker is the total time they must spend to complete all jobs assigned to them. The objective is to distribute the jobs among the workers in such a way that the maximum working time for any single worker is minimized. This means we want to find the least amount of time any worker has to work if the jobs are assigned optimally. The function should return this minimum maximum working time.

Intuition

To solve this problem, we use a backtracking (or depth-first search) approach. Since we want to minimize the maximum working time of any worker, a brute force approach would involve trying all possible assignments of jobs to workers and finding the assignment with the minimum maximum working time, which would be very inefficient. This is where backtracking becomes helpful, as it allows us to explore possibilities but backtrack as soon as we know a certain assignment cannot be better than the best one found so far.

We start by sorting the jobs in decreasing order. This ensures that we are dealing with the largest jobs first, which gives us the chance to deal with difficult cases early and potentially prune the search space more effectively.

Using the dfs function (depth-first search), we recursively explore different job assignments, where we assign the i-th job to each worker one by one and recursively assign the next job. If adding the current job to a worker's existing workload would exceed the current best answer (which is stored in the global variable ans), we skip that assignment to prune the search space.

When we assign a job to a worker whose current workload is 0 (meaning they have not been assigned a job yet), and after this assignment, their workload becomes non-zero, we don't need to try assigning the job to other workers with 0 workloads. This is because the order in which workers without any assigned jobs start their first job doesn't matter.

The ans variable keeps track of the best (minimum) maximum working time found so far, and the cnt array tracks the current working time for each worker. The algorithm terminates when all jobs have been assigned and updates the ans with the minimum of the current maximum working times if it's better than the previously recorded maximum.

Learn more about Dynamic Programming, Backtracking and Bitmask patterns.

Solution Approach

The solution provided follows the backtracking approach to solve the problem efficiently. Here's a step-by-step walkthrough of the implementation of the solution based on the reference approach provided above:

1. Sorting the jobs array: Sorting the jobs in descending order helps in handling the largest jobs first during the backtracking process. This early handling of more significant jobs can lead to earlier pruning of the branches in the search tree, hence optimizing the performance.

   1jobs.sort(reverse=True)

2. Depth-First Search (DFS): The dfs function is the core of the solution, which performs depth-first search (or backtracking).

   • When a job is to be assigned, the function iterates over all k workers to try and assign the job to each one of them.
   • It uses a counter array cnt, where cnt[j] is the total amount of time worker j has been assigned jobs.

   1def dfs(i):

3. Pruning the search space: Within the dfs function, there is a check to see if assigning the current job to a worker will result in a total working time that exceeds the best solution found so far (ans). If so, this branch of the recursion is abandoned; otherwise, the job is added to the worker's total time.

   1if cnt[j] + jobs[i] >= ans:
   2    continue

4. Exploring and backtracking: After assigning a job to a worker, the recursion continues with the next job (i + 1).

   1cnt[j] += jobs[i]
   2dfs(i + 1)
   3cnt[j] -= jobs[i]  # Backtrack

5. Avoiding identical workers' permutations: If a worker has not yet been assigned any job (cnt[j] == 0), there's no need to continue to the next worker after assigning a job to them because all workers are identical.

   1if cnt[j] == 0:
   2    break

6. Updating the answer: Once all jobs have been considered (i == len(jobs)), the function updates the minimum maximum working time found so far (ans) if the current assignment is better.

   1ans = min(ans, max(cnt))

7. Initialization: Before diving into the backtracking, initialize the search with initial values.

   1cnt = [0] * k
   2ans = inf

8. Kick off the DFS: Start the DFS with the first job (i=0).

   1dfs(0)

9. Return the result: After the backtracking is complete, the ans variable holds the desired result, which is the minimum maximum working time across all workers.

   1return ans

The algorithm makes use of recursive backtracking and pruning techniques to navigate the space of possible allocations, aiming to minimize the maximum working time of any worker. The key here is to explore different combinations of job assignments in a depth-first search manner while trimming branches that exceed the current best solution, which efficiently leads to the minimum possible maximum working time.

Example Walkthrough

Let's use a small example to illustrate the solution approach. Suppose we have the following input:

• jobs = [5,3,2,1]
• k = 2 workers available

Now let's walk through the backtracking solution:

1. Sorting the jobs array: We sort the jobs array in descending order to deal with the larger jobs first.

   1jobs.sort(reverse=True) # jobs = [5, 3, 2, 1]

2. Depth-First Search (DFS): Initialize the dfs function and the global variables cnt and ans.

   1cnt = [0] * k # cnt = [0, 0]
   2ans = float('inf')

3. First DFS call

   • We start our DFS from the first job; dfs(0) means we are considering how to assign the job with a time requirement of 5.

4. Exploring job assignments: We try assigning the job to each worker.

   • Assign job 5 to worker 0. cnt = [5, 0], then make a recursive DFS call dfs(1) and go to the next job.
   • Assign job 3 to worker 0, cnt = [8, 0], then dfs(2). Since the sum is less than ans, we continue.
   • Assign job 2 to worker 0, cnt = [10, 0], then dfs(3). Since the sum exceeds the best so far, we backtrack.
   • Assign job 2 to worker 1, cnt = [8, 2], then dfs(3). We continue because it's still less than ans.
   • Assign job 1 to worker 0, cnt = [9, 2], then we check if all jobs are assigned & update ans to 9.
   • Backtrack and assign job 1 to worker 1, cnt = [8, 3], update ans to 8.
   • Backtrack to when job 3 was assigned to worker 0.
   • Try assigning job 3 to worker 1 instead, cnt = [5, 3], and proceed with the next jobs similarly.

5. Backtracking and pruning: The DFS algorithm will backtrack whenever it encounters a sum that cannot be better than the current ans, which is constantly being updated.

6. Skip identical workers' states: When a job is assigned to a worker with a 0 workload, we don't have to consider permutations with other workers with 0 workload.

After running through all possible combinations and backtracking when necessary, we find that the best distribution that minimizes the maximum working time for a worker is [5, 3] and [2, 1] with a maximum working time of 8. So the function will return 8.

This example demonstrates the depth-first search combined with pruning to avoid unnecessary work, leading us to the solution efficiently.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the minimumTimeRequired function is determined by the depth-first search (DFS) as it tries to assign jobs to workers in every possible combination to find the minimum possible time.

• Since jobs are being sorted initially, this contributes O(n log n) to the time complexity, where n is the number of jobs.
• The DFS process has a worst-case time complexity of O(k^n) where k is the number of workers, and n is the number of jobs, since each job can be assigned to any of the k workers.
• However, due to pruning:
  • If cnt[j] + jobs[i] >= ans, the DFS does not continue down that path, which can significantly reduce the number of explored states.
  • If cnt[j] == 0, it breaks the inner loop, avoiding redundant assignments to empty slots which are equivalent.

The exact time complexity is hard to characterize due to these pruning strategies, but the worst-case without pruning is O(k^n), factoring in the sorting of jobs, we have O(n log n + k^n).

Space Complexity

For space complexity, the following considerations are taken into account:

• The recursive call stack of DFS contributes O(n) space complexity since the maximum depth of the recursion stack is n (the number of jobs).
• An array cnt of size k is used to keep track of the current sum of job times for each worker, contributing O(k) space complexity.

Therefore, the overall space complexity is O(n + k) due to the recursive stack and the cnt array. However, if we consider that the depth of the recursive stack dominates as n can potentially be larger than k, the space complexity simplifies to O(n).

Learn more about how to find time and space complexity quickly using problem constraints.