How do you solve Find Minimum Time to Finish All Jobs on LeetCode?

Find Minimum Time to Finish All Jobs (LeetCode #1723) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(k * 2^n) time complexity and O(k) space complexity. Key insight: Sorting jobs helps prioritize larger tasks, minimizing the maximum time.

What is the brute force solution for Find Minimum Time to Finish All Jobs?

The brute force approach for Find Minimum Time to Finish All Jobs is Brute Force, with O(n!) time complexity and O(n) space complexity. The brute force approach involves generating all possible assignments of jobs to workers and calculating the maximum working time for each assignment. This ensures we explore every possibility, but it can be very slow as the number of jobs increases. The optimal Optimal Solution approach improves this to O(k * 2^n).

What are the interview tips for Find Minimum Time to Finish All Jobs?

Explain your thought process clearly when discussing the brute force approach. Emphasize the importance of optimizing your solution and how you can reduce the search space. Practice backtracking problems to become comfortable with recursive solutions.

What are common mistakes when solving Find Minimum Time to Finish All Jobs?

Not considering the case where k > jobs.length. Failing to optimize by sorting jobs before assignment.

#1723

Find Minimum Time to Finish All Jobs

Hard

ArrayDynamic ProgrammingBacktrackingBit ManipulationBitmaskBacktrackingBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(k * 2^n)
Space	O(n)	O(k)

💡

Intuition

Time O(k * 2^n)Space O(k)

The optimal solution uses a backtracking approach combined with bit manipulation to efficiently explore job assignments. This reduces the number of combinations we need to check, allowing us to find the minimum maximum working time more quickly.

⚙️

Algorithm

3 steps

1Step 1: Sort the jobs in descending order to prioritize larger jobs first.
2Step 2: Use a backtracking function to assign jobs to workers, tracking the current maximum working time.
3Step 3: If the current maximum exceeds the best found so far, backtrack. If all jobs are assigned, update the best maximum if it's lower.

solution.py18 lines

1def minimumTimeRequired(jobs, k):
2    jobs.sort(reverse=True)
3    workers = [0] * k
4    best = float('inf')
5
6    def backtrack(i):
7        nonlocal best
8        if i == len(jobs):
9            best = min(best, max(workers))
10            return
11        for j in range(k):
12            workers[j] += jobs[i]
13            if workers[j] < best:
14                backtrack(i + 1)
15            workers[j] -= jobs[i]
16
17    backtrack(0)
18    return best

ℹ

Complexity note: This complexity arises from the backtracking approach, where we explore combinations of job assignments while keeping track of the maximum time for each worker.

1Sorting jobs helps prioritize larger tasks, minimizing the maximum time.
2Backtracking allows for efficient exploration of job assignments.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.