2425. Bitwise XOR of All Pairings - Detailed Explanation

Problem Statement

You are given two integer arrays, arr1 and arr2. Consider all pairs formed by taking one element from arr1 and one element from arr2. For each pair, compute the bitwise XOR of the two numbers. Your task is to return the XOR of all these pairwise XOR values.

In other words, if you define:

  result = (arr1[0] XOR arr2[0]) XOR (arr1[0] XOR arr2[1]) XOR ... XOR (arr1[i] XOR arr2[j]) ...
for all valid indices i and j,

you must compute and return result.

Example 1

• Input:
  • arr1 = [1, 3, 5]
  • arr2 = [4, 6]
• Explanation:
  The pairs and their XOR values are:
  • (1, 4) → 1 XOR 4 = 5
  • (1, 6) → 1 XOR 6 = 7
  • (3, 4) → 3 XOR 4 = 7
  • (3, 6) → 3 XOR 6 = 5
  • (5, 4) → 5 XOR 4 = 1
  • (5, 6) → 5 XOR 6 = 3
    Now, the XOR of all these values is:
    5 XOR 7 XOR 7 XOR 5 XOR 1 XOR 3 = 1
• Output: 1

Example 2

• Input:
  • arr1 = [2, 2]
  • arr2 = [9, 10]
• Explanation:
  The pairs and their XOR values are:
  • (2, 9) → 2 XOR 9 = 11
  • (2, 10) → 2 XOR 10 = 8
  • (2, 9) → 2 XOR 9 = 11
  • (2, 10) → 2 XOR 10 = 8
    Now, XORing these:
    11 XOR 8 XOR 11 XOR 8 = 0
• Output: 0

Example 3

• Input:
  • arr1 = [5]
  • arr2 = [1, 2, 3]
• Explanation:
  The pairs and their XOR values are:
  • (5, 1) → 5 XOR 1 = 4
  • (5, 2) → 5 XOR 2 = 7
  • (5, 3) → 5 XOR 3 = 6
    Now, XORing these:
    4 XOR 7 XOR 6 = 1
• Output: 1

Constraints

• 1 ≤ arr1.length, arr2.length ≤ 10⁵
• 0 ≤ arr1[i], arr2[j] ≤ 10⁹

Hints

1. Brute Force Insight:

   • One could loop through every possible pair, compute the XOR, and then XOR all these results together. However, with up to 10⁵ elements in each array, this approach results in up to 10¹⁰ operations, which is not feasible.

2. Observation about XOR and Repetition:

   • Notice that XOR is both commutative and associative. Also, for any number x, the property x XOR x = 0 holds.
   • Each element in arr1 will be paired with every element in arr2, and vice versa.

3. Counting Frequency of Occurrence:

   • If an element appears an even number of times in the overall XOR calculation, it cancels out to 0.
   • Therefore, focus on the parity (odd or even) of the number of times each element appears in the final computation.

Approach 1: Brute Force (Not Feasible)

Explanation

• Idea:
  Loop through every element in arr1 and every element in arr2, compute the XOR for each pair, and combine all these XORs.

• Steps:

  1. Initialize a variable result to 0.
  2. For each element a in arr1:
     • For each element b in arr2:
       • Compute a XOR b.
       • XOR this value into result.
  3. Return result.

• Drawback:
  The time complexity is O(n * m), which is not feasible for the given constraints.

Approach 2: Mathematical Insight and Optimized Solution

Explanation

Consider the final result:

  result = XOR over all pairs (a XOR b)

Each pair's XOR can be thought of as contributing the bits of a and b. Instead of directly computing every pair, notice the following:

• Each element a in arr1 appears in a pair with every element in arr2. Thus, a appears exactly |arr2| times in the overall XOR computation.
• Similarly, each element b in arr2 appears exactly |arr1| times.

Recall the property of XOR:

• XORing a number with itself an even number of times yields 0.
• XORing a number an odd number of times yields the number itself.

Thus, if |arr2| is even, every a from arr1 will contribute 0 to the final XOR; if |arr2| is odd, the overall contribution from arr1 will be the XOR of all elements in arr1.
Similarly, if |arr1| is odd, the overall contribution from arr2 will be the XOR of all elements in arr2; otherwise, if |arr1| is even, their contribution is 0.

Final Formula:

  result = (if |arr2| is odd then XOR(arr1) else 0) XOR (if |arr1| is odd then XOR(arr2) else 0)

Why This Works

• The XOR operation “cancels out” any number that appears an even number of times.
• This property lets us avoid iterating over all pairs, reducing the complexity to just two passes over the arrays.

Code Solutions

Python Code