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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 * 10`

^{4}`-2`

^{31}<= nums[i] <= 2^{31}- 1- Each integer in
`nums`

will appear twice, only two integers will appear once.

```
class Solution
{
public:
vector<int> singleNumber(vector<int>& nums)
{
// Pass 1 :
// Get the XOR of the two numbers we need to find
int diff = accumulate(nums.begin(), nums.end(), 0, bit_xor<int>());
// Get its last set bit
diff &= -diff;
// Pass 2 :
vector<int> rets = {0, 0}; // this vector stores the two numbers we will return
for (int num : nums)
{
if ((num & diff) == 0) // the bit is not set
{
rets[0] ^= num;
}
else // the bit is set
{
rets[1] ^= num;
}
}
return rets;
}
};
```

```
public class Solution {
public int[] singleNumber(int[] nums) {
// Pass 1 :
// Get the XOR of the two numbers we need to find
int diff = 0;
for (int num : nums) {
diff ^= num;
}
// Get its last set bit
diff &= -diff;
// Pass 2 :
int[] rets = {0, 0}; // this array stores the two numbers we will return
for (int num : nums)
{
if ((num & diff) == 0) // the bit is not set
{
rets[0] ^= num;
}
else // the bit is set
{
rets[1] ^= num;
}
}
return rets;
}
}
```

```
class Solution:
def singleNumber(self, nums):
s = reduce(xor, nums)
nz = s & (s-1) ^ s
num1 = reduce(xor, filter(lambda n: n & nz, nums))
return(num1, s ^ num1)
```

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