2425. Bitwise XOR of All Pairings

MediumBit ManipulationBrainteaserArray
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Problem Description

In the given problem, we have two arrays nums1 and nums2 made up of non-negative integers. We need to create a virtual array nums3, which would contain the results of performing the bitwise XOR operation on every possible pair formed by taking one number from nums1 and another from nums2. However, instead of generating nums3 explicitly, we are asked to directly calculate the bitwise XOR of all the elements it would contain.

In simpler terms, imagine we have, for example, nums1 = [1, 2] and nums2 = [3, 4]. The pairs and the bitwise XOR would be as follows:

  • 1 XOR 3
  • 1 XOR 4
  • 2 XOR 3
  • 2 XOR 4

We suspect nums3 would be [1 XOR 3, 1 XOR 4, 2 XOR 3, 2 XOR 4]. Our task is to find the cumulative XOR of these results, i.e., (1 XOR 3) XOR (1 XOR 4) XOR (2 XOR 3) XOR (2 XOR 4), without actually generating the intermediate array nums3.

Intuition

To understand the solution, let us first consider a property of the XOR operation: XORing the same number twice cancels it out, resulting in zero.

Now, let's analyze when a number in one of the arrays (nums1 or nums2) would be XORed an even or odd number of times:

  • If nums2 has an even number of elements, each element in nums1 is XORed with the elements in nums2 an even number of times, which means that their overall effect will be zero (since any number XORed with itself an even number of times is zero).
  • If nums2 has an odd number of elements, each element in nums1 is XORed with the elements in nums2 an odd number of times, which means the numbers in nums1 should be considered once in the final XOR calculation.
  • The same logic applies when considering the elements in nums2 with respect to the even or odd count of elements in nums1.

So, if the length of one of the arrays is odd, we need to consider each element in the other array for the final XOR calculation. If the lengths of both arrays are even, none of the elements are considered since the even-odd pairing would cancel out their effect.

With this intuition, we arrive at the solution approach: we initially set an answer variable ans to 0. We then check if the length of nums2 is odd and XOR all elements in nums1 to ans. Then we check if the length of nums1 is odd and XOR all elements in nums2 to ans. The final value of ans is the XOR of all elements that would be in nums3.

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Solution Approach

The solution uses a simple but clever observation about the XOR operation and its properties:

  • XOR of a number with itself: a XOR a = 0
  • XOR with zero: a XOR 0 = a
  • Commutative property: a XOR b = b XOR a
  • Associative property: a XOR (b XOR c) = (a XOR b) XOR c

These properties mean that when we XOR an even number of the same numbers, the result is 0, and when we do it an odd number of times, we get the number itself.

Let's go over the implementation details based on the solution provided:

  1. We initialize a variable ans to 0. This variable will serve as the accumulator for the XOR operations.
  2. We then check the length of nums2 using the bitwise & operator with 1 to determine if it's odd (len(nums2) & 1). The & 1 trick checks the least significant bit of a number, which represents its odd/even status (odd numbers have a least significant bit of 1, even numbers have a 0).
  3. If nums2 has an odd length, we iterate over each value in nums1 and apply the XOR operation to our ans variable. Due to the asymmetric nature of the required pairings (every element in nums1 is paired with every element in nums2), the cumulative effect when the count is odd is that the numbers in nums1 indeed contribute to the final result.
  4. We do the same for nums1 by checking if it has an odd length, iterating over nums2, and updating ans accordingly.
  5. At this stage, ans will hold the bitwise XOR of all integers that would be in nums3, and it's returned as the solution.

In the implementation, no extra data structures are needed because the solution leverages the XOR operation's properties to avoid constructing the nums3 array. This approach is notably efficient in both time and space complexity, as it requires iterating over each array only once (O(n) where n is the length of the longer array) and uses only a constant amount of extra space for the ans variable.

Another key aspect is the use of bitwise operations (^ for XOR and & for AND), which are low-level operations that are generally very fast in execution compared to higher-level arithmetic or logic operations.

No complex patterns or algorithms are used; the simplicity of the approach comes from a deep understanding of the XOR operation, which makes it a very elegant solution.

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Example Walkthrough

Let's take two arrays nums1 = [5, 9] and nums2 = [7, 11, 12] to illustrate the solution approach.

  1. We start by initializing ans to 0.
  2. We then check if the length of nums2 is odd. Since it has 3 elements, which is odd, we proceed to the next step.
  3. We iterate over each element in nums1 and XOR it with ans.
    • After XORing 5 with ans (initially 0), ans becomes 5 (5 XOR 0 = 5).
    • Next, we XOR 9 (the second element in nums1) with ans, resulting in ans = 5 XOR 9 = 12.
  4. We don't need to check the length of nums1 because we already know nums2 has an odd length, and that's enough to determine that all elements of nums1 should be part of the final XOR. If nums1 has an even length, XORing it with an even count of nums2 would cancel out its elements.

At this point, we have ans = 12, which is the cumulative XOR of array nums1 with each element in nums2 considering the odd-even pairing effect.

Since the implementation leverages the XOR operation's properties to avoid constructing nums3, the calculation is direct and efficient. The odd-length check using len(nums2) & 1 ensures that we only iterate through nums1 or nums2 if necessary.

In this example, we're avoiding the unnecessary and memory-intensive task of generating all possible pairs and their XOR results by directly accumulating the XORs that would result from such pairs. Thus, applying the algorithm's insight provides us with an answer: the bitwise XOR of all the elements that would be in the virtual array nums3 is 12.

Solution Implementation

1from typing import List
2
3class Solution:
4    def xor_all_nums(self, nums1: List[int], nums2: List[int]) -> int:
5        # Initialize the result of XOR operation to 0.
6        result_xor = 0
7      
8        # If the length of nums2 is odd, XOR all elements in nums1 with result_xor.
9        if len(nums2) % 2:
10            for num in nums1:
11                result_xor ^= num
12      
13        # If the length of nums1 is odd, XOR all elements in nums2 with result_xor.
14        if len(nums1) % 2:
15            for num in nums2:
16                result_xor ^= num
17              
18        # Return the final result after performing all XOR operations.
19        return result_xor
20
1// A class to find a solution for the XOR problem
2class Solution {
3    // Method to calculate the XOR of all elements after performing XOR as if each element of nums1 is paired with all elements of nums2 and vice-versa
4    public int xorAllNums(int[] nums1, int[] nums2) {
5        int result = 0; // Initialize result to zero
6
7        // If the length of nums2 is odd, XOR result with all elements in nums1
8        // Because if nums2 has an odd number of elements, each element in nums1 will be represented an odd number of times when taking the XOR with all elements of nums2 and vice versa.
9        if (nums2.length % 2 == 1) {
10            for (int value : nums1) {
11                result ^= value;
12            }
13        }
14
15        // Likewise, if the length of nums1 is odd, XOR result with all elements in nums2
16        if (nums1.length % 2 == 1) {
17            for (int value : nums2) {
18                result ^= value;
19            }
20        }
21
22        return result; // Return the accumulated XOR result
23    }
24}
25
1#include <vector>
2
3class Solution {
4public:
5    // Function to calculate the XOR of all elements as if they were all combined
6    // Parameter nums1 is the first vector of integers
7    // Parameter nums2 is the second vector of integers
8    // Returns the XOR of all elements combined
9    int xorAllNums(vector<int>& nums1, vector<int>& nums2) {
10        int result = 0;
11
12        // If nums2 has an odd number of elements, XOR all elements of nums1
13        if (nums2.size() % 2 == 1) {
14            for (int num : nums1) {
15                result ^= num;
16            }
17        }
18
19        // If nums1 has an odd number of elements, XOR all elements of nums2
20        if (nums1.size() % 2 == 1) {
21            for (int num : nums2) {
22                result ^= num;
23            }
24        }
25
26        // Return the final result of the XOR operation
27        return result;
28    }
29};
30
1/**
2 * XOR all numbers from two arrays in a specific way.
3 * 
4 * If one array has an odd length, XOR all numbers from the other array.
5 * If both arrays have odd lengths, XOR all numbers from both arrays.
6 * 
7 * @param {number[]} nums1 - The first array of numbers.
8 * @param {number[]} nums2 - The second array of numbers.
9 * @return {number} - The resulting XOR from the above rule.
10 */
11function xorAllNums(nums1: number[], nums2: number[]): number {
12    // Initialize the answer variable to store the final result.
13    let result = 0;
14
15    // Check if the length of nums2 is odd.
16    // If it is, accumulate the XOR of all elements in nums1 with the result.
17    if (nums2.length % 2 !== 0) {
18        result ^= nums1.reduce((accumulator, currentValue) => accumulator ^ currentValue, 0);
19    }
20  
21    // Check if the length of nums1 is odd.
22    // If it is, accumulate the XOR of all elements in nums2 with the result.
23    if (nums1.length % 2 !== 0) {
24        result ^= nums2.reduce((accumulator, currentValue) => accumulator ^ currentValue, 0);
25    }
26  
27    // Return the final XOR result.
28    return result;
29}
30
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Time and Space Complexity

Time Complexity

The time complexity of the provided code depends on the lengths of the input lists nums1 and nums2.

  1. Checking if len(nums2) & 1: This is an O(1) operation as it involves checking the parity of the length of nums2.

  2. Iterating over nums1: If the length of nums2 is odd, we iterate over all elements in nums1, resulting in O(n) complexity where n is the length of nums1.

  3. Checking if len(nums1) & 1: This is another O(1) operation similar to the first check but for nums1.

  4. Iterating over nums2: If the length of nums1 is odd, we iterate over all elements in nums2, resulting in O(m) complexity where m is the length of nums2.

Combining these operations, in the worst case, both nums1 and nums2 lengths are odd, which results in iterating over both lists. Hence, the total time complexity is O(n + m).

Space Complexity

The space complexity is related to the amount of extra space required that is not part of the input. For the given code, we only use an extra variable ans to store the intermediate results of the XOR operation, regardless of the input size.

The space complexity is thus O(1) because the space used does not scale with the size of the input.

Learn more about how to find time and space complexity quickly using problem constraints.

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