Write a query to identify duplicate rows in a Transactions table based on customer_id, amount, and date.
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SQL
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Answer
Use ROW_NUMBER() partitioned by the columns that define a "duplicate" - any partition with more than one row contains duplicates.
Solution
SELECT *
FROM (
SELECT t.*,
ROW_NUMBER() OVER (PARTITION BY customer_id, amount, date ORDER BY transaction_id) AS rn
FROM Transactions t
) x
WHERE rn > 1;Step by Step
PARTITION BY customer_id, amount, dategroups rows that share the exact same customer, amount and date together.ROW_NUMBER()assigns 1, 2, 3... within each group, ordered arbitrarily bytransaction_id(any tiebreaker column works).- Rows where
rn > 1are the "extra" copies beyond the first - the actual duplicates.
Alternative Method: GROUP BY with HAVING
SELECT customer_id, amount, date, COUNT(*) AS duplicate_count
FROM Transactions
GROUP BY customer_id, amount, date
HAVING COUNT(*) > 1;This version returns one summary row per duplicate group with a count, rather than every individual duplicate row - useful when you only need to know which combinations repeat, not each specific row.
A Common Follow-Up: Removing the Duplicates
DELETE FROM Transactions
WHERE transaction_id IN (
SELECT transaction_id FROM (
SELECT transaction_id,
ROW_NUMBER() OVER (PARTITION BY customer_id, amount, date ORDER BY transaction_id) AS rn
FROM Transactions
) x WHERE rn > 1
);Key Exam Traps
- Defining "duplicate" correctly: The interviewer's exact definition of a duplicate (which columns must match) changes the
PARTITION BYlist entirely - always confirm this first rather than assuming all columns must match. - GROUP BY alone can't identify individual rows - it only tells you duplicates exist, not which specific row to keep or delete. That's exactly why
ROW_NUMBER()is needed for row-level actions. - Ordering inside the window matters when deleting - without a deterministic
ORDER BY(like a timestamp or ID), which specific "duplicate" row survives is technically undefined behavior in some engines.