Use a monotonic stack to efficiently find the next smaller or equal element for each item, enabling O(n) time. Compute the total discount by summing these next smaller-or-equal values, then subtract from the total price to get the amount paid. Items with no such element receive no discount and should be identified.
Pro tip: Clarify whether the discount is applied per item or cumulatively, and confirm that 'next cheaper-or-equal' means the nearest subsequent element with value <= current. This shows attention to detail and avoids misinterpretation.
Restate the discount rule and confirm edge cases: equal prices count, no discount if no such element, and discounts are per item. Ask about input constraints (e.g., array size, price range).
Use a monotonic stack to find the next smaller-or-equal element for each item in O(n). Iterate through the array, maintaining a stack of indices with non-decreasing prices; when a smaller-or-equal price is found, pop and record the discount.
While processing, accumulate the total discount and track items that never get a discount (i.e., remain in the stack at the end). Alternatively, after computing discounts, sum them and collect indices with zero discount.
Walk through a small example (e.g., [10, 5, 8, 3]) to ensure the algorithm correctly computes discounts and identifies no-discount items. Check edge cases like strictly increasing or decreasing arrays.
Explain that each element is pushed and popped at most once, giving O(n) time and O(n) space for the stack. Discuss potential optimizations if space is a concern.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.