2425. Bitwise XOR of All Pairings

Method 1

You are given two 0-indexed arrays, nums1 and nums2, consisting of non-negative integers. There exists another array, nums3, which contains the bitwise XOR of all pairings of integers between nums1 and nums2 (every integer in nums1 is paired with every integer in nums2 exactly once).
Return the bitwise XOR of all integers in nums3.

Example 1

Input: nums1 = [2,1,3], nums2 = [10,2,5,0]
Output: 13
Explanation:
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.

Example 2

Input: nums1 = [1,2], nums2 = [3,4]
Output: 0
Explanation:
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.

Method 1

【O(n + m) time | O(1) space】

package Leetcode.Math;

/**
 * @author zhengxingxing
 * @date 2025/01/16
 */
public class BitwiseXOROfAllPairings {
  public static int xorAllNums(int[] nums1, int[] nums2) {
    // Initialize result variable to store the final XOR value
    int result = 0;

    /*
     * First Check: nums2.length % 2 == 1
     * Why this check?
     * - Each number in nums1 will appear nums2.length times in the final calculation
     * - If nums2.length is odd, each number in nums1 will appear odd times
     * - When a number appears odd times in XOR operations, it needs to be XORed once
     * - When a number appears even times, it cancels out (equals 0 in XOR)
     *
     * Example: If nums2.length is 3
     * - Each number in nums1 will appear 3 times in the pairs
     * - 3 times XOR is equivalent to XOR once
     * - Therefore, we need to XOR each number in nums1 once
     */
    if (nums2.length % 2 == 1) {
      for (int num : nums1) {
        result ^= num;  // XOR each number from nums1 once
      }
    }

    /*
     * Second Check: nums1.length % 2 == 1
     * Why this check?
     * - Each number in nums2 will appear nums1.length times in the final calculation
     * - If nums1.length is odd, each number in nums2 will appear odd times
     * - When a number appears odd times in XOR operations, it needs to be XORed once
     * - When a number appears even times, it cancels out (equals 0 in XOR)
     *
     * Example: If nums1.length is 3
     * - Each number in nums2 will appear 3 times in the pairs
     * - 3 times XOR is equivalent to XOR once
     * - Therefore, we need to XOR each number in nums2 once
     */
    if (nums1.length % 2 == 1) {
      for (int num : nums2) {
        result ^= num;  // XOR each number from nums2 once
      }
    }

    // Return the final XOR result
    return result;
  }

  /**
   * Main method to test the solution with various test cases
   */
  public static void main(String[] args) {
    // Test Case 1: nums1 has odd length, nums2 has even length
    int[] nums1_1 = {2, 1, 3};
    int[] nums2_1 = {10, 2, 5, 0};
    System.out.println("Test Case 1 Result: " + xorAllNums(nums1_1, nums2_1)); // Expected: 13

    // Test Case 2: Both arrays have even length
    int[] nums1_2 = {1, 2};
    int[] nums2_2 = {3, 4};
    System.out.println("Test Case 2 Result: " + xorAllNums(nums1_2, nums2_2)); // Expected: 0

    // Test Case 3: Both arrays have odd length
    int[] nums1_3 = {1};
    int[] nums2_3 = {2};
    System.out.println("Test Case 3 Result: " + xorAllNums(nums1_3, nums2_3)); // Expected: 3
  }
}

Method 1

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