Count paid orders per customer
Given `orders` with columns `order_id`, `customer`, `amount`, `status` and `placed`, return per `customer` the count of paid orders as `paid` and the count of all their orders as `total`. Every customer appears, even with zero paid. Sort by `customer` ascending.
Implement
order_mix(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[{"paid":3,"total":3,"customer":"ada"},{"paid":0,"total":2,"customer":"bo"},{"paid":1,"total":1,"customer":"cy"}]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 12 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}]}][{"paid":3,"total":3,"customer":"ada"},{"paid":0,"total":2,"customer":"bo"},{"paid":1,"total":1,"customer":"cy"}]not run yetsample