How do you solve Bitwise XOR of All Pairings on LeetCode?

Bitwise XOR of All Pairings (LeetCode #2425) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(n + m) space complexity. Key insight: The XOR operation is commutative and associative, meaning the order of operations does not matter.

What is the brute force solution for Bitwise XOR of All Pairings?

The brute force approach for Bitwise XOR of All Pairings is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves calculating the XOR for every possible pairing of elements from nums1 and nums2. This is straightforward but inefficient for larger arrays since it requires nested loops. The optimal Optimal Solution approach improves this to O(n + m).

What algorithm or data structure is used to solve Bitwise XOR of All Pairings?

Bitwise XOR of All Pairings uses the following concepts: Array, Bit Manipulation, Brainteaser. The recommended patterns to study are: Bit Manipulation, Counting Frequencies.

What are the interview tips for Bitwise XOR of All Pairings?

Always clarify the problem and constraints before jumping into coding. Think about the properties of operations (like XOR) that can simplify calculations. Discuss your thought process and approach with the interviewer to demonstrate your understanding.

What are common mistakes when solving Bitwise XOR of All Pairings?

Students often forget to account for the parity (even/odd) of counts when calculating the final XOR. Some may try to optimize prematurely without understanding the properties of XOR.

#2425

Bitwise XOR of All Pairings

Medium

ArrayBit ManipulationBrainteaserBit ManipulationCounting Frequencies

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + m)
Space	O(n)	O(n + m)

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Intuition

Time O(n + m)Space O(n + m)

Instead of generating all pairings, we can leverage the properties of XOR and the counts of each number. Each number in nums1 contributes to the final result based on how many times it pairs with numbers in nums2 and vice versa.

⚙️

Algorithm

3 steps

1Step 1: Count the occurrences of each number in nums1 and nums2.
2Step 2: For each unique number in nums1, calculate its contribution to the final XOR based on its count in nums2 and vice versa.
3Step 3: XOR all contributions together to get the final result.

solution.py16 lines

1# Full working Python code
2from collections import Counter
3
4def xor_all_pairings(nums1, nums2):
5    count1 = Counter(nums1)
6    count2 = Counter(nums2)
7    result = 0
8
9    for num in count1:
10        for num2 in count2:
11            if (count1[num] * count2[num2]) % 2 == 1:
12                result ^= (num ^ num2)
13    return result
14
15# Example usage
16print(xor_all_pairings([2, 1, 3], [10, 2, 5, 0]))

ℹ

Complexity note: The time complexity is O(n + m) because we iterate through both arrays to count occurrences and then through the unique keys in the counts. The space complexity is O(n + m) for storing the counts.

1The XOR operation is commutative and associative, meaning the order of operations does not matter.
2Each number's contribution to the final XOR depends on how many times it is paired with other numbers.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.