1235. Maximum Profit in Job Scheduling - Detailed Explanation

Notes on the Java implementation: We sort an array of jobs by end time. We then use Arrays.binarySearch on the endTimes array to find the insertion point for start_i. If start_i is not exactly found, binarySearch returns -(insertion_point) - 1. We convert that to get the index of the last job that ends before start_i. Finally, we compute dp[i] similar to the Python approach. The main method tests the implementation with the sample cases.

Complexity Analysis

• Brute Force Approach: Time Complexity: Exponential, roughly O(2^n) in the worst case. Each job can be either taken or not, leading to 2^n combinations (though conflicts reduce some combinations, the growth is still exponential). This is infeasible for large n (even n=30 would be over a billion possibilities). Space Complexity: O(n) for recursion stack in the worst case (one branch might take n nested calls). No substantial additional storage aside from recursion overhead.

• Optimized DP + Binary Search Approach: Time Complexity: O(n log n). Sorting takes O(n log n). Building the DP uses a loop of n iterations, each doing a O(log n) binary search. For n = 50,000, n log n is about 50k * 15 ≈ 750k operations, which is very efficient. Space Complexity: O(n) for the job list, end-times array, and DP array. The recursion stack is eliminated in the iterative version (or bounded by O(n) in a memoized recursion). This easily fits in memory for n=50k.

Comparison: The DP approach is exponentially faster than brute force for large inputs. A brute force might work for very small n (like n < 20) for educational purposes, but DP is the only viable approach for n up to 50,000.

Step-by-Step Example Walkthrough

Let's walk through a smaller example in detail to solidify understanding. Consider:

startTime = [1, 2, 3]  
endTime   = [3, 4, 5]  
profit    = [50, 20, 30]

Here we have 3 jobs:

• Job A: (1,3) profit 50
• Job B: (2,4) profit 20
• Job C: (3,5) profit 30

By inspection, jobs A and C do not overlap (A ends at 3, C starts at 3), so we could take A + C for profit 80. Job B overlaps with both A and C (B runs 2–4, which conflicts with A's 1–3 and C's 3–5), so B cannot go with either A or C. The best solution should be A + C = 80.

Using DP approach:

1. Sort by end time:
   Sorted order: A (end=3), B (end=4), C (end=5).

2. Initialize dp[0]=0.

3. Evaluate Job A (end=3, start=1, profit=50):

   • No earlier job ends <= 1, so j=0.
   • Include profit = 50 + dp[0] = 50.
   • Exclude profit = dp[0] = 0.
   • dp[1] = max(50, 0) = 50.

4. Evaluate Job B (end=4, start=2, profit=20):

   • No earlier job ends <= 2 (Job A ends at 3, which is >2), so j=0.
   • Include profit = 20 + dp[0] = 20.
   • Exclude profit = dp[1] = 50 (we already have 50 from job A).
   • dp[2] = max(20, 50) = 50.
     (It’s better to skip B and keep the profit from A so far.)

5. Evaluate Job C (end=5, start=3, profit=30):

   • Find last job ending <= 3. Job A ends at 3 (which counts, since start=3 is allowed), so j=1 (Job A's index in sorted order).
   • Include profit = 30 + dp[1] = 30 + 50 = 80.
   • Exclude profit = dp[2] = 50.
   • dp[3] = max(80, 50) = 80.

After processing all jobs, dp[3] = 80, which is the maximum profit (from taking jobs A and C). This matches our manual analysis. The DP table values were:

• dp[0]=0
• dp[1]=50 (take A)
• dp[2]=50 (skip B, carry over A’s profit)
• dp[3]=80 (take C in addition to A)

This step-by-step illustrates how at each job we consider the best of taking it (plus whatever was optimal before its start) or skipping it. The binary search ensured we quickly found that Job A was the right predecessor for Job C.

Common Mistakes and Tips

• Forgetting to Sort or Wrong Sort Order: Not sorting the jobs properly is a common bug. Always sort by start time for the recursive approach or by end time for the iterative DP approach. If unsorted, the logic to find the next job or previous compatible job will fail.

• Greedy Assumption: A naive greedy strategy (like always picking the job with highest profit or earliest finishing time next) can fail. Counterexample: one job with huge duration and profit vs many small jobs with slightly less total profit – the greedy pick of highest profit might lock out the optimal combination. The DP is needed to evaluate combinations properly.

• Incorrect Binary Search Bounds: When using binary search to find the next or previous compatible job, off-by-one errors are easy to make. Ensure that if a job starts exactly when another ends, you treat them as non-overlapping. For example, using bisect_right (in Python) or adjusting the index from Arrays.binarySearch (in Java) is necessary to handle equality correctly (so that end time == next start is allowed).

• Not Using Memoization: If implementing the top-down recursive solution, forgetting to memoize results will lead to recomputation of subproblems. Always cache dfs(i) results for each index i to avoid exponential time. In Python, you might use functools.lru_cache for memoization, or a dictionary/array. In Java, use an array for memo. Without memoization, the recursive solution will time out.

• Stack Overflow in Recursion: A purely recursive solution (even with memo) might hit recursion depth limits for n=50,000 (Python’s default recursion limit will be exceeded, and Java could run out of stack for very deep recursion). If using recursion, consider tail-recursion optimization (not available in Python) or switching to an iterative DP approach to be safe.

• Misinterpreting “no overlap”: Remember that if a job ends at time X and another starts at time X, this is not an overlap (they are back-to-back). A common mistake is to treat this as overlapping. Make sure your condition (startTime[j] >= endTime[i]) is using >= and not > when finding the next job.

• Using O(n^2) DP: One might attempt a DP where for each job you loop backwards to find a compatible job. That would work correctly but be too slow for large n (50k^2 operations is ~2.5 billion). Using binary search to find the compatible job index cuts this down significantly. Avoid the double nested loop approach for large inputs.

Edge Cases to Consider

• Single Job: If there’s only one job, the answer is just that job’s profit (trivial case). Make sure the code handles n=1 properly (it should, in both approaches).

• All Jobs Disjoint: If no two jobs overlap at all (each job starts after the previous one ends), then the optimal solution is to take all jobs. The algorithm should handle this and sum up all profits. (DP would just add profit each time since each job is compatible with the previous.)

• All Jobs Overlap: If every job overlaps with every other (for example, all start at the same time or all have overlapping intervals), then you can only take one job. The solution should pick the single job with the highest profit. Ensure the algorithm indeed does that (the DP would compare taking each job versus another and end up with the max single profit).

• Jobs With Same Start or Same End Times: If multiple jobs start at the same time or end at the same time, the algorithm should still work. Sorting by end time (or start) will group them, but the DP/transitions still correctly evaluate taking one versus another. Just be careful with binary search on equal times (as mentioned in mistakes). For example, two jobs with same end time – the order between them in sorting doesn’t matter for correctness, as neither can be taken with the other anyway.

• Large Time Values: Start and end times can be large (up to 1e9), but our algorithm’s complexity doesn’t depend on the magnitude of time values, only on ordering. Just ensure using appropriate data types (int is fine for 1e9 in Java, and Python ints handle it automatically).

• Large n: The solution should handle 50,000 jobs. Ensure that data structures (like recursion stack or loops) are efficient. Our DP approach uses O(n) memory and O(n log n) time which is fine. Test boundary scenarios if possible (like 50k jobs with minimal overlaps vs maximal overlaps) to ensure performance.

Alternative Approaches and Related Problems

• Top-Down Memoization: As discussed, you can implement the DP using recursion + memo (depth-first search with caching) instead of an explicit table. The strategy and complexity remain the same (O(n log n)). Some languages might find this easier to implement using built-in libraries (e.g., Python’s functools.lru_cache). The bottom-up and top-down approaches are just two ways to implement the same DP recurrence.

• Greedy for Unweighted Scheduling: If all profits were equal (or if the goal was to maximize number of jobs instead of profit), the problem reduces to the classic Activity Selection Problem. In that case, a greedy algorithm (always take the job which finishes first) would maximize the number of non-overlapping jobs. This is a simpler variant when profit doesn't matter.

• Job Sequencing with Deadlines: A related problem (often seen in interviews/competitions) is where each job has a deadline and profit, and each job takes the same unit time. The task is to schedule jobs such that they finish before their deadlines to earn profit. That can be solved with a greedy approach using a min-heap or union-find, and is different from interval scheduling because the "length" of each job is equal and you schedule in discrete slots up to the deadline.

• Weighted Interval Scheduling in General: The problem we solved is essentially Weighted Interval Scheduling. In algorithm textbooks, the same DP approach is used. For further practice, you can look at other formulations of weighted scheduling:

  • Selecting compatible intervals with maximum total weight.
  • Variants where instead of profit, you maximize something like total length of intervals (which would effectively be similar approach).
  • Problems where intervals might be defined by different criteria (e.g., tasks you can do in a day with start-end times).

• LeetCode Variations: Some other LeetCode problems to practice that involve interval scheduling or DP on intervals:

  • Non-overlapping Intervals (LeetCode 435): You need to remove the minimum number of intervals to avoid overlap (or equivalently, select the maximum number of non-overlapping intervals). This is the unweighted version solved by a greedy strategy.
  • Maximum Length of Pair Chain (LeetCode 646): Similar to activity selection, but phrased with pairs, asking for the longest chain of pairs where each chain link is compatible (can be solved greedily by end times as well).
  • Course Schedule III (LeetCode 630): A different spin on scheduling where each "job" has a duration and a deadline, and you want to take as many courses as possible before deadlines. The strategy there is greedy with a priority queue (not DP), but it’s another kind of scheduling optimization.
  • Job Scheduling with Resources: Some problems add more complexity like needing machines or having cooldowns (e.g., "Task Scheduler" problem where tasks have a cooling interval between same tasks). Those require different strategies (greedy or greedy+counting).