Achieving top N results and 'Other' category in MySQL - mysql

Question

I'm trying to get the top 10 currencies (and their request count) from a table in an MySQL database and group all other currencies in an 'Other' category with request count equal to the sum of these currencies' request counts.
The below query gives me the correct result, but is probably highly inefficient.
SET #row_number = 0;
SELECT
rank,
CASE WHEN rank <=10 THEN ccy ELSE 'Other' END as ccy,
SUM(req_count) AS requests
FROM (SELECT CASE WHEN rank <= 10 THEN rank ELSE 11 END AS rank, ccy, req_count
FROM (SELECT (#row_number:=#row_number + 1) AS rank, ccy, req_count
FROM (SELECT have_currency AS ccy, COUNT(*) AS req_count
FROM db1.table1
GROUP BY have_currency
ORDER BY req_count DESC)
AS currencies)
AS currencies)
AS currencies
GROUP BY rank ASC;
Result:
# rank, ccy, requests
'1', 'SGD', '184481'
'2', 'USD', '10723'
'3', 'MYR', '8044'
'4', 'HKD', '7316'
'5', 'THB', '5725'
'6', 'JPY', '4930'
'7', 'INR', '2767'
'8', 'AUD', '2164'
'9', 'VND', '2130'
'10', 'CNY', '1965'
'11', 'Other', '10217'
Any way to make this more efficient?
Bonus question: is it possible to return the % of total for request counts instead of the absolute number?

Answer 1

is this what you looking for ?
SELECT rank, IF(rank>10,'other',ccy) AS ccy,sum(req_count)
FROM (
SELECT #rank := (#rank := #rank+1) AS rank, ccy,req_count
FROM (
SELECT have_currency AS ccy, COUNT(*) AS req_count
FROM table1
GROUP BY have_currency
ORDER BY req_count DESC
) AS d1
) AS d2
CROSS JOIN (SELECT #rank := 0) AS param
GROUP BY LEAST(rank, 11);