Algorithm to find two repeated numbers in an array, without sorting

Question

There is an array of size n (numbers are between 0 and n - 3) and only 2 numbers are repeated. Elements are placed randomly in the array.

E.g. in {2, 3, 6, 1, 5, 4

Answer 1

I know the question is very old but I suddenly hit it and I think I have an interesting answer to it. We know this is a brainteaser and a trivial solution (i.e. HashMap, Sort, etc) no matter how good they are would be boring.

As the numbers are integers, they have constant bit size (i.e. 32). Let us assume we are working with 4 bit integers right now. We look for A and B which are the duplicate numbers.

We need 4 buckets, each for one bit. Each bucket contains numbers which its specific bit is 1. For example bucket 1 gets 2, 3, 4, 7, ...:

Bucket 0 : Sum ( x where: x & 2 power 0 == 0 )
...
Bucket i : Sum ( x where: x & 2 power i == 0 )

We know what would be the sum of each bucket if there was no duplicate. I consider this as prior knowledge.

Once above buckets are generated, a bunch of them would have values more than expected. By constructing the number from buckets we will have (A OR B for your information).

We can calculate (A XOR B) as follows:

A XOR B = Array[i] XOR Array[i-1] XOR ... 0, XOR n-3 XOR n-2  ... XOR 0

Now going back to buckets, we know exactly which buckets have both our numbers and which ones have only one (from the XOR bit).

For the buckets that have only one number we can extract the number num = (sum - expected sum of bucket). However, we should be good only if we can find one of the duplicate numbers so if we have at least one bit in A XOR B, we've got the answer.

But what if A XOR B is zero? Well this case is only possible if both duplicate numbers are the same number, which then our number is the answer of A OR B.