single_number 

 1from functools import reduce
 2
 3
 4# @leet start
 5class Solution:
 6    def singleNumber(self, nums: list[int]) -> int:
 7        """
 8        In an array where each number is present twice except for one number,
 9        this function finds that number.
10        The way it does this is by using xor.
11        If you xor a number by itself, you get 0.
12        So all of the numbers that show up twice reduce to 0, and the only number
13        that shows up once remains.
14        """
15        return reduce(lambda x, y: x ^ y, nums)
16
17
18# @leet end
19
20
21def test():
22    assert 2 + 2 == 4
class Solution: 
 6class Solution:
 7    def singleNumber(self, nums: list[int]) -> int:
 8        """
 9        In an array where each number is present twice except for one number,
10        this function finds that number.
11        The way it does this is by using xor.
12        If you xor a number by itself, you get 0.
13        So all of the numbers that show up twice reduce to 0, and the only number
14        that shows up once remains.
15        """
16        return reduce(lambda x, y: x ^ y, nums)
def singleNumber(self, nums: list[int]) -> int: 
 7    def singleNumber(self, nums: list[int]) -> int:
 8        """
 9        In an array where each number is present twice except for one number,
10        this function finds that number.
11        The way it does this is by using xor.
12        If you xor a number by itself, you get 0.
13        So all of the numbers that show up twice reduce to 0, and the only number
14        that shows up once remains.
15        """
16        return reduce(lambda x, y: x ^ y, nums)

In an array where each number is present twice except for one number, this function finds that number. The way it does this is by using xor. If you xor a number by itself, you get 0. So all of the numbers that show up twice reduce to 0, and the only number that shows up once remains.

def test(): 
22def test():
23    assert 2 + 2 == 4