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Two-part return value tripped me up more than the algorithm itself.
Clarify the problem requirements and edge cases, then propose an efficient solution using a monotonic stack to find the next smaller or equal element for each price. Explain how to compute the discounted prices and total revenue, and identify indices that sell at full price.
Pro tip: Emphasize the O(n) time complexity of the monotonic stack approach and discuss potential pitfalls like handling equal prices correctly (using <=) and edge cases such as empty arrays or single elements.
Restate the problem in your own words and confirm details: for each item, find the first item to its right with price <= current price; discount is the difference; if none, sell at full price. Return total revenue and list of indices with full price.
Mention that a brute force O(n^2) solution checks each element's right side, but an O(n) solution using a monotonic stack is optimal. Explain that the stack maintains indices of items waiting for a smaller or equal price to their right.
Iterate through prices; while stack is not empty and current price <= price at stack top, pop and record the discount for that index. Push current index. After iteration, remaining indices in stack have no discount and sell at full price.
For each item, if a discount is found, add (price - discount) to total revenue; otherwise, add full price and record the index. Return total revenue and the list of indices.
State time complexity O(n) and space O(n). Discuss edge cases: empty array, single element, all increasing prices, all decreasing prices, and duplicate prices.
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