Find duplicate element in array in time O(n)

Question

I have been asked this question in a job interview and I have been wondering about the right answer.

You have an array of numbers from 0 to n-1, one o

Answer 1

This is an alternative solution in O(n) time and O(1) space. It is similar to rici's. I find it a bit easier to understand but, in practice, it will overflow faster.

Let X be the missing number and R be the repeated number.

1. We can assume the numbers are from [1..n], i.e. zero does not appear. In fact, while looping through the array, we can test if zero was found and return immediately if not.

2. Now consider:

   sum(A) = n (n + 1) / 2 - X + R
   
   product(A) = n! R / X
   

where product(A) is the product of all element in A skipping the zero. We have two equations in two unknowns from which X and R can be derived algebraically.

Edit: by popular demand, here is a worked-out example:

Let's set:

S = sum(A) - n (n + 1) / 2
P = n! / product(A)

Then our equations become:

R - X = S
X = R P

which can be solved to:

R = S / (1 - P)
X = P R = P S / (1 - P)

Example:

A = [0 1 2 2 4]

n = A.length - 1 = 4
S = (1 + 2 + 2 + 4) - 4 * 5 / 2 = -1
P = 4! / (1 * 2 * 2 * 4) = 3 / 2

R = -1 / (1 - 3/2) = -1 / -1/2 = 2
X = 3/2 * 2 = 3