Maximum Profit in Job Scheduling

Maximum Profit in Job Scheduling

1. What is this problem about?

The Maximum Profit in Job Scheduling interview question is a high-level optimization problem. You are given several jobs, each with a start time, end time, and a profit. You want to pick a set of non-overlapping jobs to maximize your total profit.

Two jobs overlap if the start time of one is less than the end time of another. This is a classic "Weighted Interval Scheduling" problem, which cannot be solved with a simple greedy approach.

2. Why is this asked in interviews?

This Maximum Profit in Job Scheduling coding problem is a heavy-hitter asked by elite firms like Databricks, Citadel, and Google. It tests your ability to combine multiple algorithmic techniques: Sorting, Dynamic Programming, and Binary Search. It's a comprehensive test of a candidate's architectural thinking—how to structure a complex solution that remains efficient ($O(N \log N)$) despite the potential for $O(N^2)$ overlaps.

3. Algorithmic pattern used

The Array, Sorting, Binary Search, Dynamic Programming interview pattern is the standard approach.

1. Sort all jobs based on their end times. This is crucial for defining subproblems.
2. Define dp[i] as the maximum profit using a subset of the first i jobs.
3. For each job i:
   • Choice 1: Don't include job i. Profit = dp[i-1].
   • Choice 2: Include job i. Profit = job[i].profit + dp[latest_non_overlapping_job].
4. Use Binary Search to find the latest_non_overlapping_job (the job that ends before or at job[i].start).

4. Example explanation

Jobs:

• A: [1, 3, 50]
• B: [2, 4, 10]
• C: [3, 5, 40]
• D: [3, 6, 70]

Sorted by end time: A, B, C, D.

1. dp[0] (Job A): 50
2. dp[1] (Job B): max(dp[0], B.profit + dp[none]) = max(50, 10) = 50.
3. dp[2] (Job C): max(dp[1], C.profit + dp[A]) = max(50, 40 + 50) = 90.
4. dp[3] (Job D): max(dp[2], D.profit + dp[A]) = max(90, 70 + 50) = 120.

Max profit is 120.

5. Common mistakes candidates make

A major error in the Maximum Profit in Job Scheduling coding problem is sorting by start time instead of end time. While start-time sorting can work with different DP logic, end-time sorting is much more intuitive for "previous state" lookups. Another mistake is using a linear search instead of binary search to find the previous job, which degrades the complexity to $O(N^2)$. Finally, forgetting to handle the base case (when no previous job exists) can lead to index out-of-bounds errors.

6. Interview preparation tip

Master "DP + Binary Search." It's a powerful combination for any problem where you need to look up a value from a sorted history of states. Also, practice converting a list of objects or arrays into a structured format that's easy to sort and search.