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September 1, 2023 00:57
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Find repeating number runs of length N
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IF OBJECT_ID('tempdb..#gaps','U') IS NOT NULL DROP TABLE #gaps; --SELECT * FROM #gaps | |
CREATE TABLE #gaps ( | |
ID int NOT NULL IDENTITY, | |
Val int NOT NULL, | |
); | |
DECLARE @repeat int, @randval int, @sql nvarchar(MAX); | |
DECLARE @c int = 1000; -- How many islands to create | |
WHILE (@c > 0) | |
BEGIN; | |
SELECT @repeat = FLOOR(RAND(CHECKSUM(NEWID()))*(20))+1 -- How many consecutive rows to insert | |
, @randval = FLOOR(RAND(CHECKSUM(NEWID()))*(100))+1 -- What random value to use | |
SELECT @sql = CONCAT('INSERT INTO #gaps (Val) VALUES ','(',@randval,')', REPLICATE(CONCAT(',(',@randval,')'), @repeat-1)) | |
EXEC sys.sp_executesql @sql; | |
SET @c -= 1; | |
END; | |
------------------------------------------------------------------------------ | |
GO | |
------------------------------------------------------------------------------ | |
-- Find islands of size N | |
DECLARE @n int = 10 | |
SELECT x.IslandID, IslandStart = MIN(x.ID), IslandEnd = MAX(x.ID), Val = MAX(x.Val), IslandSize = COUNT(*) | |
FROM ( | |
SELECT x.ID, x.Val | |
, IslandID = SUM(x.IslandStart) OVER (ORDER BY x.ID) | |
FROM ( | |
SELECT g.ID, g.Val | |
, IslandStart = IIF(g.Val <> LAG(g.Val, 1, 0) OVER (ORDER BY g.ID), 1, 0) | |
FROM #gaps g | |
) x | |
) x | |
GROUP BY x.IslandID | |
HAVING COUNT(*) = @n | |
ORDER BY x.IslandID |
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