Took me a second to realize the naive approach of just multiplying the last k elements each time is fine for small inputs but they clearly wanted something better.
Start by clarifying the requirements: the data type of values, the range of k, and whether k can exceed the current list size. Then propose a solution using a dynamic array (or list) to store elements and maintain a running product of the last k elements, updating it efficiently on each addition. Discuss trade-offs between time and space, and consider edge cases like k=0 or negative values.
Pro tip: Mention that you can optimize for O(1) time per operation by maintaining a product of the last k elements, but be prepared to handle division by zero if zeros are allowed. Alternatively, use a sliding window with a queue and recompute product when needed, showing awareness of the zero-handling trade-off.
Ask about the data type of values (integers, floats?), the range of k, whether k can be larger than the list size, and if negative numbers or zeros are allowed. This ensures you design the correct solution.
Decide on a dynamic array (like Python list or Java ArrayList) to store elements, and possibly a queue or circular buffer to maintain the last k elements. Consider if you need to track the product separately.
Append the new value to the list. Update the product of the last k elements: if the list size exceeds k, remove the oldest element's contribution (if using division) or recompute the product from the last k elements.
Return the precomputed product if maintained, or compute the product of the last k elements on the fly. Handle edge cases: if k > list size, return product of all elements; if k=0, return 1 (empty product).
Discuss the trade-offs: maintaining a running product gives O(1) add and O(1) getProduct, but requires handling zeros. Recomputing on each getProduct gives O(k) time but simpler code. Space is O(n) for storing elements.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.