Minimum time to finish jobs
Detailed guide and Python implementation for the 'Minimum time to finish jobs' problem.
1. Concept Overview
The 'Minimum time to finish jobs' problem is a key challenge in the Greedy section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Minimum time to finish jobs.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Minimum time to finish jobs carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Write a function min_time_jobs(jobs, k, t) that finds the minimum time to finish all jobs. The array jobs represents the time required to complete each job. There are k assignees, and each assignee takes t units of time to complete 1 unit of job. Jobs can only be assigned as contiguous sub-segments to the assignees.
- •1 <= len(jobs) <= 10^5
- •1 <= k <= len(jobs)
- •1 <= t <= 1000
- •1 <= jobs[i] <= 10^4
Examples
min_time_jobs([10, 7, 8, 12, 8, 5, 9], 4, 5)
100
Optimal contiguous assignment: [10, 7], [8, 12], [8, 5], [9]. Max job units assigned is 20 (8+12). Time = 20 * 5 = 100.
min_time_jobs([4, 5, 10], 2, 1)
10
Optimal contiguous assignment: [4, 5], [10]. Max job units assigned is 10. Time = 10 * 1 = 10.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
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Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def min_time_opt(jobs, k):
def can_finish(limit):
workers = [0] * k
def solve(i):
if i == len(jobs): return True
for j in range(k):
if workers[j] + jobs[i] <= limit:
workers[j] += jobs[i]
if solve(i + 1): return True
workers[j] -= jobs[i]
if workers[j] == 0: break
return False
return solve(0)
jobs.sort(reverse=True)
l, r = max(jobs), sum(jobs); res = r
while l <= r:
m = (l + r) // 2
if can_finish(m): res = m; r = m - 1
else: l = m + 1
return resBrute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def min_time_brute(jobs, k):
res = float('inf')
def solve(i, workers):
nonlocal res
if i == len(jobs): res = min(res, max(workers)); return
for j in range(k):
workers[j] += jobs[i]; solve(i + 1, workers); workers[j] -= jobs[i]
solve(0, [0]*k); return resAlgorithm Pattern Checklist
When dealing with Greedy data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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