Sync LeetCode submission Runtime - 1 ms (83.81%), Memory - 36.4 MB (61.27%)

Author: jimit105

2533-bitwise-xor-of-all-pairings/README.md

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+<p>You are given two <strong>0-indexed</strong> arrays, <code>nums1</code> and <code>nums2</code>, consisting of non-negative integers. There exists another array, <code>nums3</code>, which contains the bitwise XOR of <strong>all pairings</strong> of integers between <code>nums1</code> and <code>nums2</code> (every integer in <code>nums1</code> is paired with every integer in <code>nums2</code> <strong>exactly once</strong>).</p>
+
+<p>Return<em> the <strong>bitwise XOR</strong> of all integers in </em><code>nums3</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums1 = [2,1,3], nums2 = [10,2,5,0]
+<strong>Output:</strong> 13
+<strong>Explanation:</strong>
+A possible nums3 array is [8,0,7,2,11,3,4,1,9,1,6,3].
+The bitwise XOR of all these numbers is 13, so we return 13.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums1 = [1,2], nums2 = [3,4]
+<strong>Output:</strong> 0
+<strong>Explanation:</strong>
+All possible pairs of bitwise XORs are nums1[0] ^ nums2[0], nums1[0] ^ nums2[1], nums1[1] ^ nums2[0],
+and nums1[1] ^ nums2[1].
+Thus, one possible nums3 array is [2,5,1,6].
+2 ^ 5 ^ 1 ^ 6 = 0, so we return 0.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= nums1[i], nums2[j] &lt;= 10<sup>9</sup></code></li>
+</ul>

2533-bitwise-xor-of-all-pairings/solution.py

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+# Approach 2: Space Optimized Bit Manipulation
+
+# n, m = len(nums1), len(nums2)
+# Time: O(n + m)
+# Space: O(1)
+
+class Solution:
+    def xorAllNums(self, nums1: List[int], nums2: List[int]) -> int:
+        # Initialize XOR results for both arrays
+        xor1, xor2 = 0, 0
+
+        len1, len2 = len(nums1), len(nums2)
+
+        # If nums2 length is odd, each element in nums1 appears odd times in final result
+        if len2 % 2:
+            for num in nums1:
+                xor1 ^= num
+
+        # If nums1 length is odd, each element in nums2 appears odd times in final result
+        if len1 % 2:
+            for num in nums2:
+                xor2 ^= num
+
+        return xor1 ^ xor2