Identify consecutive numbers in a column (at least 3 consecutive).

To find at least 3 consecutive numbers: use gaps-and-islands via row numbering. WITH t AS (SELECT num, num - ROW_NUMBER() OVER (ORDER BY num) AS grp FROM numbers), grouped AS (SELECT grp, MIN(num) start_num, MAX(num) end_num, COUNT() cnt FROM t GROUP BY grp HAVING COUNT() >= 3) SELECT start_num, end_num, cnt FROM grouped. Logic: num - ROW_NUMBER() is constant for consecutive numbers. Alternative with LAG/LEAD: SELECT num FROM (SELECT num, LAG(num) OVER (ORDER BY num) prev, LEAD(num) OVER (ORDER BY num) nxt FROM numbers) t WHERE num - prev = 1 AND nxt - num = 1 (finds middle of 3); for variable length use the grp approach. Why it matters: Design choices compound at scale—wrong approach can cause 100× overhead. Scalability trade-offs: Profile before optimizing; validate on sample then full. Cost implications: Suboptimal choices multiply at billion-row scale.