SQL interview question

Question

I got following question on an interview: Given a table of natural numbers with some missing ones, provide output of two tables, beginning of number gap in first table and ending in second. Example:

 ____    ________
|    |   |   |   |
| 1  |   | 3 | 3 |
| 2  |   | 6 | 7 |
| 4  |   | 10| 12|
| 5  |   |___|___|
| 8  |
| 9  |
| 13 |
|____|

Answer 1

While this is pretty much the same as Phil Sandler's answer, this should return two separate tables (and I think it looks cleaner) (it works in SQL Server, at least):

DECLARE @temp TABLE (num int)
INSERT INTO @temp VALUES (1),(2),(4),(5),(8),(9),(13)

DECLARE @min INT, @max INT
SELECT @min = MIN(num), @max = MAX(num) FROM @temp

SELECT t.num + 1 AS range_start
    FROM @temp t
    LEFT JOIN @temp t2 ON t.num + 1 = t2.num
    WHERE t.num < @max AND t2.num IS NULL

SELECT t.num - 1 AS range_end
    FROM @temp t
    LEFT JOIN @temp t2 ON t.num - 1 = t2.num
    WHERE t.num > @min AND t2.num IS NULL

Answer 2

This is SQL Server syntax:

CREATE TABLE #temp (columnA int)

INSERT INTO #temp VALUES(1)
INSERT INTO #temp VALUES(2)
INSERT INTO #temp VALUES(4)
INSERT INTO #temp VALUES(5)
INSERT INTO #temp VALUES(8)
INSERT INTO #temp VALUES(9)
INSERT INTO #temp VALUES(13)

SELECT 
    t1.columnA - 1
FROM 
    #temp t1
    LEFT JOIN #temp t2 ON t1.columnA = t2.ColumnA + 1
WHERE 
    t2.ColumnA IS NULL
    AND t1.ColumnA != (SELECT MIN(ColumnA) from #temp)  

SELECT 
    t1.columnA + 1
FROM 
    #temp t1
    LEFT JOIN #temp t2 ON t1.columnA = t2.ColumnA - 1
WHERE 
    t2.ColumnA IS NULL  
    AND t1.ColumnA != (SELECT MAX(ColumnA) from #temp)  

DROP table #temp