Removing duplicates based on one value - postgresql

Question

customer id name Pay_type
1111 aaaa regular
1111 aaaa late
1111 aaaa regular
1111 aaaa regular
2222 bbbb regular
2222 bbbb regular
2222 bbbb regular
3333 cccc regular
3333 cccc late
4444 dddd regular
4444 dddd regular
I have a SQL query that gives me the above result and I want the result to remove any customer that has a late fee
the output needs to be:
customer id name Pay_type
2222 bbbb regular
2222 bbbb regular
2222 bbbb regular
4444 dddd regular
4444 dddd regular
select
distinct a.customer_id,
a.name,
pay_type
from table a
left join table b on a.customer_id= b.id
left join table c on c.id = b.pay_id
where b.status = 'Done

Answer 1

I'd do this as an anti-join:
select *
from table a
where not exists (
select null
from table b
where
a.customer_id = b.customer_id and
b.pay_type = 'late'
)
This has advantages over a distinct or a "not in" approach in that it will stop looking after it finds a match. This should work efficiently for both large and small datasets.
Any solution that uses distinct would have to evaluate the entire dataset and then remove dupes.

Answer 2

I'm not sure exactly what your tables look like, but you could do something like:
WHERE customer_id NOT IN (
SELECT customer_id
FROM table_with_customer_and_pay_type
WHERE pay_type = 'late'
GROUP BY customer_id )

Answer 3

Common Table Expression variation:
WITH orig_result_set AS (
select
distinct a.customer_id,
a.name,
pay_type
from table a
left join table b on a.customer_id= b.id
left join table c on c.id = b.pay_id
where b.status = 'Done'
),
exclude_late_payments AS (
SELECT DISTINCT customer_id
FROM orig_result_set
WHERE pay_type = 'late'
),
on_time_payments AS (
SELECT customer_id,
name,
pay_type
FROM orig_result_set
WHERE customer_id NOT IN exclude_late_payments
)
SELECT *
FROM on_time_payments