Easy interview question got harder: given numbers 1..100, find the missing number(s) given exactly k are missing

Question

I had an interesting job interview experience a while back. The question started really easy:

Q1: We have a bag containing numbers

Answer 1

You can solve Q2 if you have the sum of both lists and the product of both lists.

(l1 is the original, l2 is the modified list)

d = sum(l1) - sum(l2)
m = mul(l1) / mul(l2)

We can optimise this since the sum of an arithmetic series is n times the average of the first and last terms:

n = len(l1)
d = (n/2)*(n+1) - sum(l2)

Now we know that (if a and b are the removed numbers):

a + b = d
a * b = m

So we can rearrange to:

a = s - b
b * (s - b) = m

And multiply out:

-b^2 + s*b = m

And rearrange so the right side is zero:

-b^2 + s*b - m = 0

Then we can solve with the quadratic formula:

b = (-s + sqrt(s^2 - (4*-1*-m)))/-2
a = s - b

Sample Python 3 code:

from functools import reduce
import operator
import math
x = list(range(1,21))
sx = (len(x)/2)*(len(x)+1)
x.remove(15)
x.remove(5)
mul = lambda l: reduce(operator.mul,l)
s = sx - sum(x)
m = mul(range(1,21)) / mul(x)
b = (-s + math.sqrt(s**2 - (-4*(-m))))/-2
a = s - b
print(a,b) #15,5

I do not know the complexity of the sqrt, reduce and sum functions so I cannot work out the complexity of this solution (if anyone does know please comment below.)