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

Using scipy lib:

import math
from scipy.optimize import fsolve

def mullist(a):
    mul = 1
    for i in a:
        mul = mul*i
    return mul

a = [1,2,3,4,5,6,9,10]
s = sum(a)
so = sum(range(1,11))
mulo = mullist(range(1,11))
mul = mullist(a)
over = mulo/mul
delta = so -s
# y = so - s -x
# xy = mulo/mul
def func(x):
    return (so -s -x)*x-over

print int(round(fsolve(func, 0))), int(round(delta - fsolve(func, 0)))

Timing it:

$ python -mtimeit -s "$(cat with_scipy.py)" 

7 8

100000000 loops, best of 3: 0.0181 usec per loop

Other option is:

>>> from sets import Set
>>> a = Set(range(1,11))
>>> b = Set([1,2,3,4,5,6,9,10])
>>> a-b
Set([8, 7])

And the timing is:

Set([8, 7])
100000000 loops, best of 3: 0.0178 usec per loop