daily

Author: Zanger67

my-submissions/m2425.md

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+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.

my-submissions/m2425.py

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+class Solution:
+    def xorAllNums(self, nums1: List[int], nums2: List[int]) -> int:
+        return (reduce(xor, map(int, nums1)) if len(nums2) % 2 else 0) ^ (reduce(xor, map(int, nums2)) if len(nums1) % 2 else 0)