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

Assuming that L is a list of integers with no duplicates, you can infer that the part of the list between start and index is completely consecutive if and only if L[index] == L[start] + (index - start) and similarly with index and end is completely consecutive if and only if L[index] == L[end] - (end - index). This combined with splitting the list into two recursively gives a sublinear solution.

# python 3.3 and up, in older versions, replace "yield from" with yield loop

def missing_elements(L, start, end):
    if end - start <= 1: 
        if L[end] - L[start] > 1:
            yield from range(L[start] + 1, L[end])
        return

    index = start + (end - start) // 2

    # is the lower half consecutive?
    consecutive_low =  L[index] == L[start] + (index - start)
    if not consecutive_low:
        yield from missing_elements(L, start, index)

    # is the upper part consecutive?
    consecutive_high =  L[index] == L[end] - (end - index)
    if not consecutive_high:
        yield from missing_elements(L, index, end)

def main():
    L = [10,11,13,14,15,16,17,18,20]
    print(list(missing_elements(L,0,len(L)-1)))
    L = range(10, 21)
    print(list(missing_elements(L,0,len(L)-1)))

main()