Single Number III [LeetCode]

Problem Description

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note:

1. The order of the result is not important. So in the above example, [5, 3] is also correct.
2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

Solution 
Again, Hash Table solution is not discussed here.
This problem is an extension of the other two problems Single Number and Single Number II. As we know that for an integer N:
0 ^ N = N
N ^ N = 0
where ^ is XOR bit operator.
If we call the two numbers that we need to find is A and B. We all know that if we XOR all the elements in the provided array, we will get A ^ B. This is a lot of information!
Now, we notice that A and B must be different at some bit at position t in their binary representations. So if we divide the set of numbers into 2 set, one is the set of all the numbers that have the same bit at position t as A, the other is the set of all numbers that have the same bit at position t as B. These 2 sub-sets have a special characteristic: all numbers appear 2 times, except 1. This bring us to the Single Number problem.

I. Java Code

public class Solution {
    public int[] singleNumber(int[] nums) {
        int A = 0;
        int B = 0;
        int AXORB = 0;
        for(int i = 0; i<nums.length; i++){
            AXORB ^= nums[i];
        }
        
        AXORB = (AXORB & (AXORB - 1)) ^ AXORB; //find the different bit
        for(int i = 0; i<nums.length; i++){
            if((AXORB & nums[i]) == 0)
                A ^= nums[i];
            else
                B ^= nums[i];
        }
        return new int[]{A, B};
    }
}