Efficient way to find missing elements in an integer sequence

Question

Suppose we have two items missing in a sequence of consecutive integers and the missing elements lie between the first and last elements. I did write a code that does accomp

Answer 1

I stumbled on this looking for a different kind of efficiency -- given a list of unique serial numbers, possibly very sparse, yield the next available serial number, without creating the entire set in memory. (Think of an inventory where items come and go frequently, but some are long-lived.)

def get_serial(string_ids, longtail=False):
  int_list = map(int, string_ids)
  int_list.sort()
  n = len(int_list)
  for i in range(0, n-1):
    nextserial = int_list[i]+1
    while nextserial < int_list[i+1]:
      yield nextserial
      nextserial+=1
  while longtail:
    nextserial+=1
    yield nextserial
[...]
def main():
  [...]
  serialgenerator = get_serial(list1, longtail=True)
  while somecondition:
    newserial = next(serialgenerator)

(Input is a list of string representations of integers, yield is an integer, so not completely generic code. longtail provides extrapolation if we run out of range.)

There's also an answer to a similar question which suggests using a bitarray for efficiently handling a large sequence of integers.

Some versions of my code used functions from itertools but I ended up abandoning that approach.