MANG Problem #35: Maximum Profit in Job Scheduling (Hard)

1. Problem Statement

Mental Model

Breaking down a complex problem into its most efficient algorithmic primitive.

We have n jobs, where every job is scheduled to be done from startTime[i] to endTime[i], obtaining a profit of profit[i].

You're given the startTime, endTime, and profit arrays, return the maximum profit you can take such that there are no two jobs in the subset with overlapping time range.

If you choose a job that ends at time X you will be able to start another job that starts at time X.

2. Approach: Sort + DP + Binary Search

This is an optimization over standard 0/1 Knapsack for intervals. For every job, we can either:

Skip it: The profit is the same as the maximum profit up to the previous job.
Take it: The profit is this job's profit + the maximum profit from non-overlapping jobs before it.

To do this efficiently:

Sort: Sort all jobs by their end time.
DP Array: dp[i] stores the max profit using a subset of the first i jobs.
Binary Search: To quickly find the last non-overlapping job before job i, we use binary search (since they are sorted by end time).

3. Java Implementation

class Job {
    int start, end, profit;
    Job(int s, int e, int p) { start = s; end = e; profit = p; }
}

public int jobScheduling(int[] startTime, int[] endTime, int[] profit) {
    int n = startTime.length;
    Job[] jobs = new Job[n];
    for (int i = 0; i < n; i++) {
        jobs[i] = new Job(startTime[i], endTime[i], profit[i]);
    }
    
    // 1. Sort by end time
    Arrays.sort(jobs, (a, b) -> a.end - b.end);
    
    // dp[i] stores max profit up to job i
    int[] dp = new int[n];
    dp[0] = jobs[0].profit;
    
    for (int i = 1; i < n; i++) {
        // Option 1: Skip current job
        int skipProfit = dp[i - 1];
        
        // Option 2: Take current job
        int takeProfit = jobs[i].profit;
        int latestNonConflict = binarySearch(jobs, i);
        if (latestNonConflict != -1) {
            takeProfit += dp[latestNonConflict];
        }
        
        dp[i] = Math.max(skipProfit, takeProfit);
    }
    
    return dp[n - 1];
}

// Find the latest job that ends <= current job's start time
private int binarySearch(Job[] jobs, int currentIndex) {
    int low = 0, high = currentIndex - 1;
    while (low <= high) {
        int mid = (low + high) / 2;
        if (jobs[mid].end <= jobs[currentIndex].start) {
            if (jobs[mid + 1].end <= jobs[currentIndex].start) {
                low = mid + 1; // Can we find a later one?
            } else {
                return mid;
            }
        } else {
            high = mid - 1;
        }
    }
    return -1;
}

4. 5-Minute "Video-Style" Walkthrough

The "Aha!" Moment: If you sort jobs by end time, the problem becomes a sequence of choices. By the time you process job i, you already know the absolute best profit you could make ending before job i.
The Fast Lookup: Without binary search, finding the compatible previous job takes $O(N)$, making the overall solution $O(N^2)$. Since jobs are sorted by end time, finding the latest end time $\le$ current start time takes $O(\log N)$.
The State Array: dp[i] monotonically increases. It means "The absolute best I can do considering jobs 0 through i".

5. Interview Discussion

Interviewer: "Can this be solved using a TreeMap?"
You: "Yes! A TreeMap<Integer, Integer> mapping endTime -> maxProfit can simplify the code. We use floorEntry(startTime) to find the best previous profit. The complexity remains $O(N \log N)$."

5. Verbal Interview Script (Staff Tier)

Interviewer: "Walk me through your optimization strategy for this problem."

You: "When approaching this type of challenge, my primary objective is to identify the underlying Monotonicity or Optimal Substructure that allow us to bypass a naive brute-force search. In my implementation of 'MANG Problem #35: Maximum Profit in Job Scheduling (Hard)', I focused on reducing the time complexity by leveraging a Dynamic Programming state transition. This allows us to handle input sizes that would typically cause a standard O(N^2) approach to fail. Furthermore, I prioritized memory efficiency by optimizing the DP state to use only a 1D array. This ensures that the application remains performant even under heavy garbage collection pressure in a high-concurrency Java environment."

6. Staff-Level Interview Follow-Ups

Once you provide the optimized solution, a senior interviewer at Google or Meta will likely push you further. Here is how to handle the most common follow-ups:

Follow-up 1: "How does this scale to a Distributed System?"

If the input data is too large to fit on a single machine (e.g., billions of records), we would move from a single-node algorithm to a MapReduce or Spark-based approach. We would shard the data based on a consistent hash of the keys and perform local aggregations before a global shuffle and merge phase, similar to the logic used in External Merge Sort.

Follow-up 2: "What are the Concurrency implications?"

In a multi-threaded Java environment, we must ensure that our state (e.g., the DP table or the frequency map) is thread-safe. While we could use synchronized blocks, a higher-performance approach would be to use AtomicVariables or ConcurrentHashMap. For problems involving shared arrays, I would consider a Work-Stealing pattern where each thread processes an independent segment of the data to minimize lock contention.

7. Performance Nuances (The Java Perspective)

Autoboxing Overhead: When using HashMap<Integer, Integer>, Java performs autoboxing which creates thousands of Integer objects on the heap. In a performance-critical system, I would use a primitive-specialized library like fastutil or Trove to use Int2IntMap, significantly reducing GC pauses.
Recursion Depth: As discussed in the code, recursive solutions are elegant but risky for deep inputs. I always ensure the recursion depth is bounded, or I rewrite the logic to be Iterative using an explicit stack on the heap to avoid StackOverflowError.

Key Takeaways

Interviewer: "Can this be solved using a TreeMap?"
You: "Yes! A TreeMap<Integer, Integer> mapping endTime -> maxProfit can simplify the code. We use floorEntry(startTime) to find the best previous profit. The complexity remains $O(N \log N)$."
Problem: Analyze Recursive Depth