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First completed : January 16, 2025
Last updated : January 16, 2025
Related Topics : Array, Bit Manipulation, Brainteaser
Acceptance Rate : 67.05 %
Logic:
Take arrays A and B. If B is of even length, then each element of A will be XORed into the total of an even number of times, negating itself. This applies to all values in the array. Thus we can skip it if B is even length.
The same applies vise versa.
Besides that, an odd number of times no matter how many will be just like if there was only 1 XOR for that value since all others would negate it. This is clear when you look at the definition of an odd integer:
$x=2y+1$ where$x$ is your odd value and$y$ is an integer.
class Solution:
def xorAllNums(self, nums1: List[int], nums2: List[int]) -> int:
return (reduce(xor, nums1) if len(nums2) % 2 else 0) ^ (reduce(xor, nums2) if len(nums1) % 2 else 0)