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Home/Questions/Python/Coding/List customers with more than 5 orders.

List customers with more than 5 orders.

Python/Codingmedium0.6 min read

Reviewed by Aditya Kumar · Last reviewed 2026-03-24

**Why HAVING vs WHERE:** WHERE filters before aggregation; HAVING after. For 'more than 5 orders,' you must aggregate first—HAVING is correct. Using WHERE with a subquery is an alternative but less clear. **Scalability:** (1) Index customer_id, order_id—enables index-only scan...

🤖 Analyze Your Answer
Frequency
Low
Asked at 1 company
Category
179
questions in Python/Coding
Difficulty Split
127E|24M|28H
in this category
Total Bank
1,863
across 7 categories
Asked at these companies
Comcast
Key Concepts Tested
joinpartitionspark

Why This Question Matters

This medium-level Python/Coding question appears frequently in data engineering interviews at companies like Comcast. While less common, it tests deeper understanding that distinguishes strong candidates. Mastering the underlying concepts (join, partition, spark) will help you answer variations of this question confidently.

How to Approach This

Break this problem into components. Identify the core trade-offs involved, then walk the interviewer through your reasoning step by step. Demonstrate awareness of edge cases and production considerations - this is what separates good answers from great ones.

Expert Answer
115 words

Why HAVING vs WHERE: WHERE filters before aggregation; HAVING after. For 'more than 5 orders,' you must aggregate first—HAVING is correct. Using WHERE with a subquery is an alternative but less clear.

Scalability: (1) Index customer_id, order_id—enables index-only scan for COUNT. (2) For billions of rows: pre-aggregate in a summary table (customer_order_counts) refreshed incrementally. (3) In Spark: groupBy + filter avoids shuffle if you can push predicate—but COUNT requires shuffle.

Cost: Full table scan on 100M orders = expensive. Partition by date, use incremental aggregation, or materialize customer-level metrics.

SELECT customer_id, COUNT(order_id) AS cnt
FROM orders o JOIN customers c ON o.customer_id = c.id
GROUP BY customer_id
HAVING COUNT(order_id) > 5
ORDER BY cnt DESC;

⚡
Pro Tip

Red Flag: Using WHERE cnt > 5 (invalid—can't reference alias in WHERE). Pro-Move: 'We maintain a customer_order_counts table with daily incremental updates—queries hit that instead of raw orders.'

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According to DataEngPrep.tech, this is one of the most frequently asked Python/Coding interview questions, reported at 1 company. DataEngPrep.tech maintains an editor-reviewed database of 1,863 data engineering interview questions across 7 categories.

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