LeetCode Problem Workspace
Find Minimum Time to Finish All Jobs
Minimize the maximum working time of k workers by optimally assigning jobs, leveraging dynamic programming and bit manipulation.
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Practice Focus
Hard · State transition dynamic programming
Answer-first summary
Minimize the maximum working time of k workers by optimally assigning jobs, leveraging dynamic programming and bit manipulation.
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The problem requires assigning jobs to k workers such that the maximum work time across all workers is minimized. This can be achieved using dynamic programming combined with bit manipulation, particularly focusing on state transitions where subsets of jobs are assigned iteratively. The key idea is to minimize the maximum workload by balancing job assignments.
Problem Statement
You are given a list of jobs, where each job has a specific duration. Your task is to assign these jobs to k workers such that each worker gets at least one job. The goal is to minimize the maximum total work time assigned to any one worker, with each job assigned to exactly one worker. The time each worker spends is the sum of the durations of the jobs assigned to them.
You must determine the minimum possible maximum time any worker could have after all jobs are assigned. This is a classic problem of job assignment that can be efficiently solved using dynamic programming with state transitions and bitmasking techniques to keep track of the assigned jobs.
Examples
Example 1
Input: jobs = [3,2,3], k = 3
Output: 3
By assigning each person one job, the maximum time is 3.
Example 2
Input: jobs = [1,2,4,7,8], k = 2
Output: 11
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
Solution Approach
Dynamic Programming with Bitmasking
To solve the problem, use dynamic programming with bitmasking. Each state represents a subset of jobs that have been assigned. Transition through states by selecting a subset of jobs, assigning them to a worker, and updating the state accordingly. The key is to minimize the maximum workload by exploring all possible assignments of jobs.
State Transition and Subset Selection
In each state, we consider assigning a subset of jobs to a worker, ensuring that we minimize the maximum workload of the workers. Use bitmasking to represent all possible subsets of jobs and update the state by transitioning to the next valid state. This ensures that we find the optimal assignment for the workers.
Backtracking for Optimization
Use backtracking to explore different job assignments and keep track of the maximum working time at each step. Backtrack when the current state does not provide a better solution, ensuring we find the optimal configuration for minimizing the maximum workload.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time and space complexity depends on the number of jobs and the number of workers. The dynamic programming solution involves iterating over all subsets of jobs, leading to a time complexity of O(2^n * n * k), where n is the number of jobs and k is the number of workers. The space complexity is O(2^n), as we store the results of all subsets of jobs.
What Interviewers Usually Probe
- Candidates should demonstrate understanding of dynamic programming with bitmasking.
- Candidates should be able to explain the concept of state transitions and how they apply to this problem.
- Candidates should discuss how backtracking can optimize the solution and how it interacts with dynamic programming.
Common Pitfalls or Variants
Common pitfalls
- Not considering all subsets of jobs, leading to suboptimal assignments.
- Overcomplicating the problem by trying to brute force the solution without using dynamic programming or backtracking.
- Incorrectly transitioning between states or using the wrong bitmask representation for job assignments.
Follow-up variants
- Increasing the number of workers (k) and ensuring the algorithm still performs efficiently.
- Using additional constraints like worker availability times or different job priorities.
- Considering cases with extremely large job times to test the algorithm’s efficiency.
FAQ
How can I minimize the maximum working time for workers in the Find Minimum Time to Finish All Jobs problem?
You can minimize the maximum working time by using dynamic programming combined with bitmasking to explore all possible job assignments while minimizing the maximum workload for each worker.
What is the role of state transitions in the solution for the Find Minimum Time to Finish All Jobs problem?
State transitions allow you to incrementally build job assignments by selecting subsets of jobs for each worker, updating the current state while keeping track of the maximum working time.
What are the most common mistakes when solving the Find Minimum Time to Finish All Jobs problem?
Common mistakes include failing to explore all subsets of jobs, misrepresenting the job assignments with incorrect bitmasks, and not properly optimizing the solution with dynamic programming or backtracking.
What is bitmasking in the context of the Find Minimum Time to Finish All Jobs problem?
Bitmasking is a technique used to represent subsets of jobs as binary numbers, where each bit indicates whether a job is assigned or not. This allows efficient tracking of job assignments in the dynamic programming solution.
How can GhostInterview assist me with the dynamic programming solution for Find Minimum Time to Finish All Jobs?
GhostInterview guides you through the dynamic programming approach, helping you implement bitmasking and state transitions, and offers insight into how to handle job assignments efficiently.
Solution
Solution 1
#### Python3
class Solution:
def minimumTimeRequired(self, jobs: List[int], k: int) -> int:
def dfs(i):
nonlocal ans
if i == len(jobs):
ans = min(ans, max(cnt))
return
for j in range(k):
if cnt[j] + jobs[i] >= ans:
continue
cnt[j] += jobs[i]
dfs(i + 1)
cnt[j] -= jobs[i]
if cnt[j] == 0:
break
cnt = [0] * k
jobs.sort(reverse=True)
ans = inf
dfs(0)
return ansContinue Topic
array
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