Problem URL: Bitwise xor of all pairings

From Boolean algebra we know that n^0 = n and n^n = 0, from which we can infer that the xor of an even count of an integer n is 0, and the xor of an odd count of an integer n is n. We also know that the ^ operator is commutative and associative.

XOR of even times of a number is zero (0)
XOR of odd times of a number is number itself.

class Solution:
    def xorAllNums(self, nums1: List[int], nums2: List[int]) -> int:
        x, y = 0, 0
        for num in nums1:
            x ^= num
        for num in nums2:
            y ^= num

        return (len(nums1) % 2 * y) ^ (len(nums2) % 2 * x)

Time complexity: O(n), n is the length of the array
Space complexity: O(1)