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

Something like this:

SELECT col1, col2 FROM
(
    SELECT x + 1 as col1, 
        ROW_NUMBER() OVER(ORDER BY x) AS 'rownum'  
    FROM tbl y 
    WHERE NOT EXISTS (SELECT x FROM tbl z WHERE z.x = y.x + 1) 
        AND x <> (SELECT MAX(x) FROM tbl)
) a
INNER JOIN
(
    SELECT x - 1 as col2,
        ROW_NUMBER() OVER(ORDER BY x) AS 'rownum'  
    FROM tbl y 
    WHERE NOT EXISTS (SELECT x FROM tbl z WHERE z.x = y.x - 1) 
        AND x <> (SELECT MIN(x) FROM tbl)
) b
ON a.rownum = b.rownum

The "rownum" syntax will be different for different DBMS. The above might work for SQL Server, but I haven't tested it.

As one of the comments pointed out, many DBMS's have analytics that will make this easier.

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

Answer 3

Itzik Ben Gan writes a lot about these "gaps and islands" problems. His row_number solution to this is

WITH C AS
(
SELECT N, ROW_NUMBER() OVER (ORDER BY N) AS RN
FROM t
)
SELECT Cur.N+1,Nxt.N-1
FROM C AS Cur 
JOIN C AS Nxt ON Nxt.RN = Cur.RN+1
WHERE Nxt.N-Cur.N>1

And a solution without row_number from the same source.

SELECT N+1 AS start_range,
(SELECT MIN(B.N) FROM t AS B WHERE B.N > A.N)-1 AS end_range
FROM t AS A
WHERE NOT EXISTS(SELECT * FROM t AS B WHERE B.N = A.N+1)
AND N< (SELECT MAX(N) FROM t)

Answer 4

This works without DB Specific SQL and it could probably be made a little cleaner but it does work

EDIT: You can see this working at on this Query at StackExchange Data Explorer

SELECT low,high FROM 

(

SELECT col1, low 

FROM
(Select n1.col1 col1, min(n2.col1) + 1 low
 from numbers n1
inner join numbers n2
on n1.col1 < n2.col1 

Group by n1.col1) t
WHERE t.low not in (SELECT col1 FROM NUMBERS)
and t.low < (Select MAX(col1) from numbers) 
) t

INNER JOIN 
(

SELECT col1 - 1 col1, high
 FROM
(Select n1.col1 col1 , min(n2.col1) - 1 high
 from numbers n1
inner join numbers n2
on n1.col1 < n2.col1 

Group by n1.col1) t
WHERE t.high not in (SELECT col1 FROM NUMBERS) 
) t2
ON t.col1 = t2.col1

Answer 5

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