## Algorithm

Problem Name: 260. 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.

## Code Examples

### #1 Code Example with C Programming

```Code - C Programming```

``````
int* singleNumber(int* nums, int numsSize, int* returnSize) {
int i, k, a, b, *p;

k = 0;
for (i = 0; i  <  numsSize; i ++) {
k ^= nums[i];
}

k = k & ~(k - 1);
a = b = 0;
for (i = 0; i  <  numsSize; i ++) {
if (nums[i] & k) {
a ^= nums[i];
} else {
b ^= nums[i];
}
}

p = malloc(2 * sizeof(int));
p[0] = a;
p[1] = b;
*returnSize = 2;
return p;
}
``````
Copy The Code &

Input

cmd
nums = [1,2,1,3,2,5]

Output

cmd
[3,5]

### #2 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public int[] singleNumber(int[] nums) {
int xorValue = Integer.highestOneBit(Arrays.stream(nums)
.boxed()
.reduce((a, b) -> a ^ b)
.orElse(0));
int[] result = new int[2];
for (int num : nums) {
if ((xorValue & num) == 0) {
result[0] ^= num;
} else {
result[1] ^= num;
}
}
return result;
}
}
``````
Copy The Code &

Input

cmd
nums = [1,2,1,3,2,5]

Output

cmd
[3,5]

### #3 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const singleNumber = function(nums) {
const hash = {};
nums.forEach((el, idx) => {
if (hash.hasOwnProperty(el)) {
hash[el] += 1;
delete hash[el];
} else {
hash[el] = 1;
}
});
return Object.keys(hash).map(el => +el);
};
``````
Copy The Code &

Input

cmd
nums = [-1,0]

Output

cmd
[-1,0]

### #4 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def singleNumber(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
return [n[0] for n in collections.Counter(nums).most_common()[-2:]]
``````
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Input

cmd
nums = [-1,0]

Output

cmd
[-1,0]

### #5 Code Example with C# Programming

```Code - C# Programming```

``````
namespace LeetCode
{
public class _0260_SingleNumberIII
{
public int[] SingleNumber(int[] nums)
{
foreach (int num in nums)

int x = 0;

foreach (int num in nums)
if ((num & diff) != 0)
x ^= num;

return new int[] { x, bitmask ^ x };
}
}
}
``````
Copy The Code &

Input

cmd
nums = [0,1]

Output

cmd
[1,0]