The technical interview included two coding-style tasks: a SQL analytics query and merging two sorted linked lists. ### Constraints & Assumptions - For SQL, return one row per customer, including customers with no orders. - Report `total_orders`, `total_amount`, `latest_order_date`, and `latest_order_amount`. - Use a nested query, CTE, or window function to identify each customer's most recent order. - For linked lists, reuse existing nodes and return the merged head. ### Clarifying Questions to Ask - If a customer has multiple orders on the same latest date, how should ties be handled? - Should returned orders affect total order amount? - Should `total_amount` be `0` or `NULL` for customers with no orders? - Are linked-list values sorted in non-decreasing order? ### Part 1 - Write The SQL Query Given `customers`, `orders`, and `returns`, write a query returning customer-level order metrics and latest order amount. #### What This Part Should Cover - Aggregate orders by customer. - Use `ROW_NUMBER()` or similar to find the latest order per customer. - `LEFT JOIN` from customers so customers without orders remain. - `COALESCE` for zero counts and amounts if desired. ### Part 2 - Merge Two Sorted Linked Lists Given heads of two sorted singly linked lists, merge them into one sorted list by reusing nodes. #### What This Part Should Cover - Dummy head and tail pointer. - Compare current node values and append the smaller node. - Attach the remaining suffix. - Complexity and edge cases. ### What a Strong Answer Covers - Preserves customers with no orders. - Handles latest-order tie assumptions explicitly. - Avoids double-counting from unnecessary joins to returns. - Reuses linked-list nodes without allocating new list nodes. ### Follow-up Questions - How would you exclude returned orders? - How would you handle ties in latest order date? - How would you merge `k` sorted linked lists? - What is the recursive linked-list solution?