Job Sequencing
Detailed guide and Python implementation for the 'Job Sequencing' problem.
1. Concept Overview
The 'Job Sequencing' 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 Job Sequencing.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Job Sequencing 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 job_scheduling(jobs) that takes a list of tuples jobs where each tuple is (job_id, deadline, profit). Each job takes 1 unit of time to complete. Find the maximum profit and the number of jobs completed, and return them as a tuple (job_count, max_profit).
- •1 <= len(jobs) <= 10^4
- •1 <= deadline <= 100
- •1 <= profit <= 1000
Examples
job_scheduling([(1, 4, 20), (2, 1, 10), (3, 1, 40), (4, 1, 30)])
(2, 60)
We can do job 3 at time 1 and job 1 at time 2. Total profit is 40 + 20 = 60.
job_scheduling([(1, 2, 100), (2, 1, 19), (3, 2, 27), (4, 1, 25), (5, 3, 15)])
(3, 142)
We can do job 1, 3, and 5 for a total profit of 100 + 27 + 15 = 142.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
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 job_scheduling_opt(jobs):
jobs.sort(key=lambda x: x[1], reverse=True)
max_d = max(j[0] for j in jobs)
slot = [-1] * (max_d + 1); count = 0; profit = 0
for j in jobs:
for s in range(j[0], 0, -1):
if slot[s] == -1:
slot[s] = 1; count += 1; profit += j[1]; break
return count, profitBrute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def job_scheduling_brute(jobs):
# Try all permutations and check deadlines (not practical)
return job_scheduling_opt(jobs)Algorithm Pattern Checklist
When dealing with Greedy data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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