## 1. Problem

Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.

You must implement a solution with a linear runtime complexity and use only constant extra space.

Example 1:

```
Input: nums = [2,2,1]
Output: 1
```

Example 2:

```
Input: nums = [4,1,2,1,2]
Output: 4
```

Example 3:

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

Constraints:

- 1 <= nums.length <= 3 * 10^4
- -3 * 10^4 <= nums[i] <= 3 * 10^4
- Each element in the array appears twice except for one element which appears only once.

## 2. Solution

I solved this problem like this.

- Complexity
- Time complexity : O(N)
- Space complexity : O(N)

- Step
- Count nums using dictionary.
- Checking value is 1 and return key.

```
class Solution:
def singleNumber(self, nums: List[int]) -> int:
dic = {}
for num in nums:
if num not in dic:
dic[num] = 0
dic[num] += 1
for key, value in dic.items():
if value == 1:
return key
return -1
```

Other solution is XOR operation. That code is very short!

```
def singleNumber(self, nums: List[int]) -> int:
return reduce(lambda total, el: total ^ el, nums)
```