Amazon SQL job interview question: customers who made 2+ purchases - sql

Question

You have a simple table that has only two fields: CustomerID, DateOfPurchase. List all customers that made at least 2 purchases in any period of six months. You may assume the table has the data for the last 10 years. Also, there is no PK or unique value.
My friend already got the job, despite the fact that he couldn't answer this question. I was curious how this kind of question can be solved.
Thanks

Answer 1

From an abstract view this problem is about efficiently self joining a table with no PK or unique identifier.
This is very tricky as you see there can be scenarios like
a customer making exactly 2 purchase in 6 month that too on same date (which can look like duplicate record)
a customer making >=2 purchase in 6 month on different date(the usual case).
One of the thing that needs to be done here is generate a column that can act
like a unique identifier which can be achieved here using row_number
After having a unique identifier it is easy to join on your required conditions and unique identifier from 1st alias != unique identifier from 2nd alias (meaning joining all rows from both alias except with same row, same row != different row with same data as in 1st scenario)
Putting it all together, it can achieved using
common table expressions to start with a single source of data that includes a manually added unique identifier and then doing the required business
row_number which helps us assign that unique identifier to our single source of data generated in a common table expression.
refer the below query for technical details.
with tempPurchase as (
select *,
row_number() over (order by CustomerID) as rowNumber -- this is crucial part
from purchase
)
select distinct(tp1.CustomerID) from tempPurchase as tp1
join tempPurchase as tp2 on tp1.CustomerID = tp2.CustomerID
and tp1.DateOfPurchase >= tp2.DateOfPurchase
and tp1.DateOfPurchase <= DATEADD(month, 6, tp2.DateOfPurchase)
and tp1.rowNumber != tp2.rowNumber; -- this is crucial part
Refer db fiddle here for complete working solution.

Answer 2

We can try using exists logic here to detect records for the same customer occurring within 6 months. Then, find distinct customers, which implies that any such matching customer has at least two purchases.
SELECT DISTINCT CustomerID
FROM yourTable t1
WHERE EXISTS (SELECT 1 FROM yourTable t2
WHERE t2.CustomerID = t1.CustomerID AND
t2.DateOfPurchase > t1.DateOfPurchase AND
t2.DateOfPurchase <= DATEADD(month, 6, t1.DateOfPurchase));
Note that this answer assumes that there would only be at most one distinct purchase per day by a given customer. A better approach would be:
SELECT DISTINCT CustomerID
FROM yourTable t1
WHERE EXISTS (SELECT 1 FROM yourTable t2
WHERE t2.CustomerID = t1.CustomerID AND
t2.PK <> t1.PK AND
t2.DateOfPurchase >= t1.DateOfPurchase AND
t2.DateOfPurchase <= DATEADD(month, 6, t1.DateOfPurchase));
The above query reads as saying find, for each customer, any relationship between 2 records within 6 months of each other which are distinct purchases. This assumes that the table has a PK primary key column. Ideally, every table should have some kind of logical primary key.

Answer 3

Try this:
SELECT distinct CustomerID
FROM purchase t1
WHERE 1 < (SELECT count(1) FROM purchase t2
WHERE t2.CustomerID = t1.CustomerID AND
t2.DateOfPurchase >= t1.DateOfPurchase AND
t2.DateOfPurchase <= DATEADD(month, 6, t1.DateOfPurchase))
Idea is to pick one record from outer table t1 and check in inner table t2 if there are any purchases made within 6 months including the one which you picked from outer table. If count from subquery is greater than 1 then we have the eligible customer.