SQL Interview Trap 🚨 Consecutive Orders Logic

You have a table:
orders

order_id | customer_id | order_date | amount

👉 Question:

Find customers who placed orders on 3 or more consecutive days,
but return only the first date of each such streak per customer.

⚠️ No temp tables.
⚠️ Assume multiple orders per day are possible.

🧠 Most candidates fail because they:

- Forget to handle multiple orders on the same day
- Misuse ROW_NUMBER()
- Miss the date gap logic

✅ Correct SQL Solution:

WITH distinct_orders AS (
SELECT DISTINCT customer_id, order_date
FROM orders
),
grp AS (
SELECT
customer_id,
order_date,
order_date - INTERVAL '1 day' *
ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY order_date) AS grp_id
FROM distinct_orders
)
SELECT
customer_id,
MIN(order_date) AS streak_start_date
FROM grp
GROUP BY customer_id, grp_id
HAVING COUNT(*) >= 3;

💡 Why this works (Interview Gold):

- DISTINCT removes same-day duplicates
- ROW_NUMBER() creates a sequence
- Date minus row number groups consecutive dates
- HAVING COUNT(*) >= 3 filters valid streaks

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