With sql find next available integer within range that is not present in existing integer subset(s)

Question

Problem statement:

given a range x -> y of unsigned integers
where x and y are both in the range

Answer 1

After a lot of thinking, I believe a query as simple as this will do:

with a as(
  -- next ip address
  select n.next_address, i.subnet_sk
  from ip_address i
  CROSS APPLY (SELECT convert(binary(4), convert(bigint, i.address) + 1) AS next_address) as n
  where n.next_address NOT IN (SELECT address FROM ip_address)
  AND EXISTS (SELECT 1 FROM subnet s WHERE s.subnet_sk = i.subnet_sk and n.next_address > s.ipv4_begin and n.next_address < s.ipv4_end)

  UNION -- use UNION here, not UNION ALL to remove duplicates

  -- first ip address for completely unassigned subnets
  SELECT next_address, subnet_sk
  FROM subnet 
  CROSS APPLY (SELECT convert(binary(4), convert(bigint, ipv4_begin) + 1) AS next_address) n
  where n.next_address NOT IN (SELECT address FROM ip_address)

  UNION -- use UNION here, not UNION ALL to remove duplicates

  -- next ip address from dhcp ranges
  SELECT next_address, subnet_sk
  FROM dhcp_range
  CROSS APPLY (SELECT convert(binary(4), convert(bigint, end_address) + 1) AS next_address) n
  where n.next_address NOT IN (SELECT address FROM ip_address)
)
SELECT min(next_address), subnet_sk
FROM a WHERE NOT exists(SELECT 1 FROM dhcp_range dhcp
         WHERE a.subnet_sk = dhcp.subnet_sk and a.next_address
            between dhcp.begin_address
                and dhcp.end_address)
GROUP BY subnet_sk

It is for IPV4, but can be easily extended for IPV6

SQLFiddle

Results for each subnet:

           subnet_sk
---------- -----------
0xAC101129 1
0xC0A81B1F 2
0xC0A8160C 3

(3 row(s) affected)

In my opinion it should be very fast. Please check it