2425. Bitwise XOR of All Pairings

[mah-shamim]

Topics: Array, Bit Manipulation, Brainteaser

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.

Constraints:

• 1 <= nums1.length, nums2.length <= 105
• 0 <= nums1[i], nums2[j] <= 109

Hint:

1. Think how the count of each individual integer affects the final answer.
2. If the length of nums1 is m and the length of nums2 is n, then each number in nums1 is repeated n times and each number in nums2 is repeated m times.

[mah-shamim]

We need to think carefully about how the bitwise XOR operates and how the pairing of elements in nums1 and nums2 affects the result.

Key Observations:

1. Each element in nums1 pairs with every element in nums2, and vice versa.

2. In nums3, each number from nums1 and nums2 appears multiple times:

   • Each number from nums1 appears n times (where n is the length of nums2).
   • Each number from nums2 appears m times (where m is the length of nums1).

3. The XOR operation has a property that if a number appears an even number of times, it will cancel out (because x ^ x = 0), and if it appears an odd number of times, it contributes to the final XOR result.

Approach:

• For each bit position (from 0 to 31, since nums1[i] and nums2[j] are at most 10^9), we will calculate how many times this bit is set (i.e., 1) in the XOR result of all pairings.
• Each number in nums1 contributes to the final XOR n times, and each number in nums2 contributes to the final XOR m times.

Using these observations, we can reduce the problem to counting the number of times each bit is set (odd number of times) across all pairings.

Plan:

1. Count the number of 1s in each bit position for nums1 and nums2.
2. For each bit, check whether the total count of 1s (from both nums1 and nums2) for that bit is odd or even.
3. If the count is odd, set that bit in the final XOR result.

Let's implement this solution in PHP: 2425. Bitwise XOR of All Pairings

<?php
function xorAllNums($nums1, $nums2) {
    $m = count($nums1); // length of nums1
    $n = count($nums2); // length of nums2
    $result = 0;
    
    // Iterate over all 32 possible bit positions (0 to 31)
    for ($bit = 0; $bit < 32; $bit++) {
        $count1 = 0; // Count of 1's in the $bit-th position in nums1
        $count2 = 0; // Count of 1's in the $bit-th position in nums2
        
        // Count 1's in nums1 for the $bit-th position
        foreach ($nums1 as $num) {
            if (($num >> $bit) & 1) {
                $count1++;
            }
        }
        
        // Count 1's in nums2 for the $bit-th position
        foreach ($nums2 as $num) {
            if (($num >> $bit) & 1) {
                $count2++;
            }
        }
        
        // Calculate the number of 1's in the $bit-th position in nums3
        $totalCount = $count1 * $n + $count2 * $m;
        
        // If totalCount is odd, set the $bit-th bit in result
        if ($totalCount % 2 == 1) {
            $result |= (1 << $bit);
        }
    }
    
    return $result;
}

// Example 1
$nums1 = [2, 1, 3];
$nums2 = [10, 2, 5, 0];
echo xorAllNums($nums1, $nums2); // Output: 13

// Example 2
$nums1 = [1, 2];
$nums2 = [3, 4];
echo xorAllNums($nums1, $nums2); // Output: 0
?>

Explanation:

• $m and $n: Store the lengths of nums1 and nums2 respectively.
• $result: This variable will store the final XOR result.
• Loop over 32 bits: Since each number is at most 10^9, we only need to consider the first 32 bits.
• Counting bits: For each bit position, we count how many times this bit is set in nums1 and nums2.
• Calculate total counts: Each bit from nums1 contributes n times to the total count, and each bit from nums2 contributes m times.
• XOR result update: If the total count for a particular bit is odd, we update the result to set that bit.

Complexity:

• Time Complexity: O(m + n), where m and n are the lengths of nums1 and nums2, respectively.
• Space Complexity: O(1), as we use constant extra space.

Example Walkthrough:

Example 1:

$nums1 = [2, 1, 3];
$nums2 = [10, 2, 5, 0];
echo xorAllPairings($nums1, $nums2);  // Output: 13

Example 2:

$nums1 = [1, 2];
$nums2 = [3, 4];
echo xorAllPairings($nums1, $nums2);  // Output: 0

This solution is efficient and works within the problem's constraints.

[basharul-siddike]

Thanks for solving the problem of "Bitwise XOR of All Pairings"

[mah-shamim]

Thanks