Bitwise XOR of All Pairings

Bitwise XOR of All Pairings - Problem

You are given two 0-indexed arrays, nums1 and nums2, consisting of non-negative integers.

Let there be 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.

Input & Output

Example 1 — Basic Case

$ Input: nums1 = [2,1], nums2 = [3,4]

› Output: 2

💡 Note: All pairs: (2,3)→1, (2,4)→6, (1,3)→2, (1,4)→5. Final XOR: 1⊕6⊕2⊕5 = 2

Example 2 — Single Elements

$ Input: nums1 = [1], nums2 = [2]

› Output: 3

💡 Note: Only one pair: (1,2) → 1⊕2 = 3

Example 3 — Same Values

$ Input: nums1 = [1,2], nums2 = [3,4]

› Output: 0

💡 Note: Pairs: (1,3)→2, (1,4)→5, (2,3)→1, (2,4)→6. XOR: 2⊕5⊕1⊕6 = 0

Constraints

1 ≤ nums1.length, nums2.length ≤ 10⁵
0 ≤ nums1[i], nums2[j] ≤ 10⁹

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is that XOR has a cancellation property: elements that appear an even number of times cancel out. Each element in nums1 appears exactly len(nums2) times in all pairs, and vice versa. The optimal approach checks if these frequencies are odd and only XORs elements with odd frequencies. Best approach is XOR Properties with Time: O(n+m), Space: O(1).

Common Approaches

✓ Brute Force - Generate All Pairs

⏱️ Time: O(n×m) Space: O(1)

Create all pairs between nums1 and nums2, compute their XOR values, then XOR all results together. This directly follows the problem description but is inefficient for large arrays.

Optimized - XOR Properties

⏱️ Time: O(n+m) Space: O(1)

Key insight: each element in nums1 appears exactly len(nums2) times in all pairs, and each element in nums2 appears exactly len(nums1) times. If an element appears an even number of times in XOR operations, it cancels out.

Brute Force - Generate All Pairs — Algorithm Steps

Iterate through each element in nums1
For each nums1 element, iterate through each element in nums2
XOR the pair and accumulate the result

Visualization

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Step-by-Step Walkthrough

Arrays

Two input arrays nums1 and nums2

Generate Pairs

Create all possible pairs and XOR each pair

Final XOR

XOR all pair results together

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums1, int nums1Size, int* nums2, int nums2Size) {
    int result = 0;
    for (int i = 0; i < nums1Size; i++) {
        for (int j = 0; j < nums2Size; j++) {
            result ^= (nums1[i] ^ nums2[j]);
        }
    }
    return result;
}

int main() {
    char line1[10000], line2[10000];
    fgets(line1, sizeof(line1), stdin);
    fgets(line2, sizeof(line2), stdin);
    
    // Remove newline
    line1[strcspn(line1, "\n")] = 0;
    line2[strcspn(line2, "\n")] = 0;
    
    int nums1[1000], nums2[1000];
    int nums1Size = 0, nums2Size = 0;
    
    // Parse first array
    char* token = strtok(line1 + 1, ",]");
    while (token != NULL) {
        nums1[nums1Size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    // Parse second array
    token = strtok(line2 + 1, ",]");
    while (token != NULL) {
        nums2[nums2Size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    printf("%d\n", solution(nums1, nums1Size, nums2, nums2Size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n×m)

Nested loops through nums1 (n elements) and nums2 (m elements)

✓ Linear Growth

Space Complexity

O(1)

Only using a single variable to accumulate the result

✓ Linear Space

23.5K Views

Medium Frequency

~15 min Avg. Time

892 Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Generate All Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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