Find "n" consecutive free numbers from table

I have some table with numbers like this (status is either FREE or ASSIGNED)

id_set number status
-----------------------
1 000001 ASSIGNED
1 000002 FREE
1 000003 ASSIGNED
1 000004 FREE
1 000005 FREE
1 000006 ASSIGNED
1 000007 ASSIGNED
1 000008 FREE
1 000009 FREE
1 000010 FREE
1 000011 ASSIGNED
1 000012 ASSIGNED
1 000013 ASSIGNED
1 000014 FREE
1 000015 ASSIGNED

and I need to find "n" consecutive numbers, so for n = 3, query would return

1 000008 FREE
1 000009 FREE
1 000010 FREE

It should return only first possible group of each id_set (in fact, it would be executed only for id_set per query)

I was checking WINDOW functions, tried some queries like COUNT(id_number) OVER (PARTITION BY id_set ROWS UNBOUNDED PRECEDING), but that's all I got :) I couldn't think of logic, how to do that in Postgres.

I was thinking about creating virtual column using WINDOW functions counting preceding rows for every number where status = 'FREE', then select first number, where count is equal to my "n" number.

Or maybe group numbers by status, but only from one ASSIGNED to another ASSIGNED and select only groups containing at least "n" numbers

EDIT

I found this query (and changed it a little bit)

WITH q AS
(
  SELECT *,
         ROW_NUMBER() OVER (PARTITION BY id_set, status ORDER BY number) AS rnd,
         ROW_NUMBER() OVER (PARTITION BY id_set ORDER BY number) AS rn
  FROM numbers
)
SELECT id_set,
       MIN(number) AS first_number,
       MAX(number) AS last_number,
       status,
       COUNT(number) AS numbers_count
FROM q
GROUP BY id_set,
         rnd - rn,
         status
ORDER BY
     first_number

which produces groups of FREE/ASSIGNED numbers, but I would like to have all numbers from only first group which meets the condition

SQL Fiddle

This is a gaps-and-islands problem. Assuming there are no gaps or duplicates in the same id_set set:

WITH partitioned AS (
  SELECT
    *,
    number - ROW_NUMBER() OVER (PARTITION BY id_set) AS grp
  FROM atable
  WHERE status = 'FREE'
),
counted AS (
  SELECT
    *,
    COUNT(*) OVER (PARTITION BY id_set, grp) AS cnt
  FROM partitioned
)
SELECT
  id_set,
  number
FROM counted
WHERE cnt >= 3
;

Here's a SQL Fiddle demo* link for this query: http://sqlfiddle.com/#!1/a2633/1.

UPDATE

To return only one set, you could add in one more round of ranking:

WITH partitioned AS (
  SELECT
    *,
    number - ROW_NUMBER() OVER (PARTITION BY id_set) AS grp
  FROM atable
  WHERE status = 'FREE'
),
counted AS (
  SELECT
    *,
    COUNT(*) OVER (PARTITION BY id_set, grp) AS cnt
  FROM partitioned
),
ranked AS (
  SELECT
    *,
    RANK() OVER (ORDER BY id_set, grp) AS rnk
  FROM counted
  WHERE cnt >= 3
)
SELECT
  id_set,
  number
FROM ranked
WHERE rnk = 1
;

Here's a demo for this one too: http://sqlfiddle.com/#!1/a2633/2.

If you ever need to make it one set per id_set, change the RANK() call like this:

RANK() OVER (PARTITION BY id_set ORDER BY grp) AS rnk

Additionally, you could make the query return the smallest matching set (i.e. first try to return the first set of exactly three consecutive numbers if it exists, otherwise four, five etc.), like this:

RANK() OVER (ORDER BY cnt, id_set, grp) AS rnk

or like this (one per id_set):

RANK() OVER (PARTITION BY id_set ORDER BY cnt, grp) AS rnk

* The SQL Fiddle demos linked in this answer use the 9.1.8 instance as the 9.2.1 one doesn't appear to be working at the moment.

StackExchange Database Administrators Q#36943, answer score: 20