Latest order per customer
Given `orders` with columns `order_id`, `customer`, `amount`, `status` and `placed`, return the latest order per `customer`, breaking ties on `placed` by the larger `order_id`. Return `customer`, `order_id` and `amount`, sorted by `customer` ascending.
Implement
latest_order(orders: dataframe) → dataframeExamples
in
[{"__df__":[{"amount":120.5,"placed":"2024-01-05","status":"paid","customer":"ada","order_id":1},{"amount":80,"placed":"2024-01-07","status":"refunded","customer":"bo","order_id":2},{"amount":45.25,"placed":"2024-02-11","status":"paid","customer":"ada","order_id":3},{"amount":200,"placed":"2024-02-14","status":"paid","customer":"cy","order_id":4},{"amount":15.75,"placed":"2024-03-02","status":"pending","customer":"bo","order_id":5},{"amount":60,"placed":"2024-03-19","status":"paid","customer":"ada","order_id":6}]}]out[{"amount":60,"customer":"ada","order_id":6},{"amount":15.75,"customer":"bo","order_id":5},{"amount":200,"customer":"cy","order_id":4}]What a strong answer looks like
State your approach and its time/space complexity out loud before you optimize. Handle the edge cases (empty input, duplicates, overflow), and say why you chose this over the brute force. Green tests are the floor, not the grade.
0:00 of about 13 min
solution.py
InputExpectedGot
[{"__df__":[{"amount":120.5,"placed":"2024-01-05","status":"paid","customer":"ada","order_id":1},{"amount":80,"placed":"2024-01-07","status":"refunded","customer":"bo","order_id":2},{"amount":45.25,"placed":"2024-02-11","status":"paid","customer":"ada","order_id":3},{"amount":200,"placed":"2024-02-14","status":"paid","customer":"cy","order_id":4},{"amount":15.75,"placed":"2024-03-02","status":"pending","customer":"bo","order_id":5},{"amount":60,"placed":"2024-03-19","status":"paid","customer":"ada","order_id":6}]}][{"amount":60,"customer":"ada","order_id":6},{"amount":15.75,"customer":"bo","order_id":5},{"amount":200,"customer":"cy","order_id":4}]not run yetsample