Top Spenders: Find the Top N Customers by Total Spend Using SQL and Python
Interview Experience
Problem
You have two tables: customers(id, name) and orders(id, customer_id, amount, created_at). Write a SQL query to find the top 5 customers by total spend, along with their order count and average order value.
sql
-- customers: id INT, name VARCHAR
-- orders: id INT, customer_id INT, amount DECIMAL, created_at TIMESTAMP
SELECT
c.name,
COUNT(o.id) AS order_count,
SUM(o.amount) AS total_spend,
AVG(o.amount) AS avg_order_value
FROM customers c
JOIN orders o ON c.id = o.customer_id
GROUP BY c.id, c.name
ORDER BY total_spend DESC
LIMIT 5;
Example output:
name | order_count | total_spend | avg_order_value
------------|-------------|-------------|----------------
Alice | 12 | 4800.00 | 400.00
Bob | 8 | 3200.00 | 400.00
Follow-ups
- How would you modify the query to show top spenders per month using a window function?
- What index would you add to
ordersto speed up this aggregation? - Rewrite this in Python using pandas from a DataFrame -- what is the equivalent
groupbychain? - How would you handle customers who have placed no orders (LEFT JOIN behavior)?
Full Details
Problem
You have two tables: customers(id, name) and orders(id, customer_id, amount, created_at). Write a SQL query to find the top 5 customers by total spend, along with their order count and average order value.
sql
-- customers: id INT, name VARCHAR
-- orders: id INT, customer_id INT, amount DECIMAL, created_at TIMESTAMP
SELECT
c.name,
COUNT(o.id) AS order_count,
SUM(o.amount) AS total_spend,
AVG(o.amount) AS avg_order_value
FROM customers c
JOIN orders o ON c.id = o.customer_id
GROUP BY c.id, c.name
ORDER BY total_spend DESC
LIMIT 5;
Example output:
name | order_count | total_spend | avg_order_value
------------|-------------|-------------|----------------
Alice | 12 | 4800.00 | 400.00
Bob | 8 | 3200.00 | 400.00
Follow-ups
- How would you modify the query to show top spenders per month using a window function?
- What index would you add to
ordersto speed up this aggregation? - Rewrite this in Python using pandas from a DataFrame -- what is the equivalent
groupbychain? - How would you handle customers who have placed no orders (LEFT JOIN behavior)?
About This Question
This is a candidate experience report from a cloudkitchens interview during the phone round.
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How To Practice This Type of Question
Solve similar problems on LeetCode under timed conditions (25-35 minutes per medium difficulty). The goal is pattern recognition: recognize the underlying technique (sliding window, two-pointer, BFS, memoized recursion, etc.) within 60-90 seconds of reading. Strong candidates verbalize their hypothesis out loud before coding, then iterate based on feedback. Weak candidates dive into implementation immediately, lose time on the wrong approach, and run out of time for follow-ups.
Companies update their question pools every 2-4 months. The exact wording of any given question may have been retired by the time you interview. Focus your prep on the pattern, not the specific problem. The patterns that appear in Cloudkitchens reports consistently are the ones worth investing in; one-off niche problems are not.
During Your Cloudkitchens Round
Apply the standard interview round template: clarify requirements (2-3 minutes), state your approach out loud and confirm direction with the interviewer (3-5 minutes), code with narration (15-25 minutes), test with concrete examples including edge cases (5 minutes), discuss optimization or trade-offs if time permits (5 minutes). This template is universally accepted across FAANG and adjacent companies; deviating from it produces weaker interviewer feedback signal.
The single most predictive failure mode in Cloudkitchens reports tagged "no hire": not asking clarifying questions. Interviewers are explicitly trained to weight this. Strong candidates ask 3-5 clarifying questions even on problems that look obvious; weak candidates dive into code immediately. The clarifying-question check is often the first signal recorded in the interviewer's written notes.