# Single Number III LeetCode Solution | Easy Approach Share:

Single Number III Given an integer array `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. You can return the answer in any order.

You must write an algorithm that runs in linear runtime complexity and uses only constant extra space.

Example 1:

```Input: nums = [1,2,1,3,2,5]
Output: [3,5]
Explanation:  [5, 3] is also a valid answer.
```

Example 2:

```Input: nums = [-1,0]
Output: [-1,0]
```

Example 3:

```Input: nums = [0,1]
Output: [1,0]
```

Constraints:

• `2 <= nums.length <= 3 * 104`
• `-231 <= nums[i] <= 231 - 1`
• Each integer in `nums` will appear twice, only two integers will appear once.

Time: O(n)
Space: O(1)

### C++

``````class Solution {
public:
vector<int> singleNumber(vector<int>& nums) {
const int xors = accumulate(begin(nums), end(nums), 0, bit_xor<>());
const int lowbit = xors & -xors;
vector<int> ans(2);

// seperate nums into two groups by the lowbit
for (const int num : nums)
if (num & lowbit)
ans ^= num;
else
ans ^= num;

return ans;
}
};
``````

### Java

`````` class Solution {
public int[] singleNumber(int[] nums) {
final int xors = Arrays.stream(nums).reduce((a, b) -> a ^ b).getAsInt();
final int lowbit = xors & -xors;
int[] ans = new int;

// seperate nums into two groups by the lowbit
for (final int num : nums)
if ((num & lowbit) > 0)
ans ^= num;
else
ans ^= num;

return ans;
}
}
``````

### Python

``````
class Solution:
def singleNumber(self, nums: List[int]) -> List[int]:
xors = reduce(operator.xor, nums)
lowbit = xors & -xors
ans = [0, 0]

# seperate nums into two groups by the lowbit
for num in nums:
if num & lowbit:
ans ^= num
else:
ans ^= num

return ans
``````

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