Maximum Profitable Triplets With Increasing Prices II

Maximum Profitable Triplets With Increasing Prices II

1. What is this problem about?

The Maximum Profitable Triplets With Increasing Prices II interview question is the more advanced version of the previous problem. It typically involves larger constraints, making even an $O(N^2)$ approach insufficient. You still need to find three indices i < j < k such that prices[i] < prices[j] < prices[k] and maximize the sum of their profits.

With N up to $10^5$, you must find an $O(N \log N)$ or $O(N \log (\max\_price))$ solution. This usually requires advanced data structures for range maximum queries.

2. Why is this asked in interviews?

IBM and other high-tech firms use the Maximum Profitable Triplets With Increasing Prices II coding problem to screen for candidates with strong competitive programming backgrounds. It tests your ability to implement and adapt complex data structures like Segment Trees or Fenwick Trees (Binary Indexed Trees). It also evaluates your understanding of coordinate compression—a vital technique when you need to use values as indices in a tree structure.

3. Algorithmic pattern used

The Array, Binary Indexed Tree, Segment Tree interview pattern is mandatory here.

1. Coordinate Compression: Since prices can be large, map them to their relative ranks (1 to N).
2. Left Max Profits: Use a Fenwick Tree to iterate from left to right. For each prices[j], query the max profit of any price smaller than prices[j] seen so far.
3. Right Max Profits: Use another Fenwick Tree to iterate from right to left. For each prices[j], query the max profit of any price larger than prices[j] seen so far.
4. For each j, combine the results: max_left_profit[j] + profits[j] + max_right_profit[j].

4. Example explanation

Prices: [1, 5, 2, 4, 3], Profits: [10, 50, 20, 40, 30]

1. Left Pass (Max profit for price < current):
   • At price 5: max(price 1) = 10.
   • At price 2: max(price 1) = 10.
   • At price 4: max(price 1, 2) = 20.
   • At price 3: max(price 1, 2) = 20.
2. Right Pass (Max profit for price > current):
   • At price 2: max(price 4, 3) = 40.
   • At price 1: max(price 5, 2, 4, 3) = 50.
3. Calculate for j=2 (Price 2): 10 (Left) + 20 (Self) + 40 (Right) = 70.
4. Calculate for j=1 (Price 5): 10 (Left) + 50 (Self) + 0 (Right) = 60.

The algorithm would find the maximum across all j.

5. Common mistakes candidates make

In the Maximum Profitable Triplets With Increasing Prices II coding problem, the most common mistake is not using coordinate compression, leading to excessive memory usage if prices are large. Another error is not correctly handling the "strictly increasing" condition (e.g., using <= instead of < in the tree query). Implementing a Fenwick Tree for "maximum" instead of the standard "sum" is also a point where candidates often trip up.

6. Interview preparation tip

Practice implementing a standard Fenwick Tree and a Segment Tree from scratch. Understand how to modify them for different types of queries (Sum, Min, Max, GCD). For Range Maximum Queries, a Fenwick Tree only works if the values you are inserting are non-decreasing over time, otherwise, a Segment Tree is more robust.