Classic monotonic stack problem once you see it, but I spent a few minutes trying to brute force it first.
Clarify the problem and edge cases, then propose an efficient solution using a monotonic stack to find the next smaller or equal element for each price in O(n) time. Walk through the algorithm with a small example, then discuss implementation details and complexity.
Pro tip: Mention that this is a classic 'Next Smaller Element' problem and that a monotonic stack is optimal; also note that handling equal prices correctly (using <=) is crucial to avoid off-by-one errors.
Confirm that for each item, we need the first subsequent item with price <= current price, and if none exists, the discount is 0 (final price unchanged).
Recognize that a monotonic stack efficiently finds the next smaller or equal element in O(n) time by maintaining a stack of indices with increasing prices.
Iterate through the array; while the stack is not empty and the current price is <= the price at the stack's top index, pop and set the discount for that index to the current price. Push the current index onto the stack.
After the iteration, any indices left in the stack have no later item with a lower or equal price, so their discount remains 0.
State that the time complexity is O(n) and space complexity is O(n) due to the stack. Walk through a small example to verify correctness, including edge cases like all increasing or all equal prices.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.