Bitwise XOR of All Pairings
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Bit Manipulation
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.
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 <= 1050 <= nums1[i], nums2[j] <= 109
Solution
Clarifying Questions
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
- What is the range of values for the integers within the input array?
- Can the input array be empty or null?
- Are there any constraints on the size of the input array?
- Are duplicate numbers allowed in the input array, and if so, how should they be handled?
- What is the expected behavior if the input array contains only a single element?
Brute Force Solution
Approach
The brute force approach to this problem is pretty straightforward. It means that we consider every single pair of numbers from the given list, perform the XOR operation on each pair, and then combine all those XOR results using XOR again.
Here's how the algorithm would work step-by-step:
- Take the first number in the list and find the XOR with every other number, one at a time.
- Now take the second number in the list, and do the same thing: XOR it with every other number, including the first.
- Keep doing this for every single number in the list, making sure you find the XOR of that number with every other number.
- Each time you calculate the XOR of a pair, remember to combine it with your running total. You can do this with XOR as well.
- At the very end, the final total is the answer you're looking for.
Code Implementation
def bitwise_xor_of_all_pairings(numbers):
xor_total = 0
# Iterate through each number in the input list
for first_number_index in range(len(numbers)):
for second_number_index in range(len(numbers)):
# XOR the two numbers at the current indices
xor_result = numbers[first_number_index] ^ numbers[second_number_index]
# Accumulate the XOR result into the running total
# This combines the XOR of each pair
xor_total ^= xor_result
return xor_totalBig(O) Analysis
Time Complexity
O(n²) – The provided algorithm iterates through the input list of size n. For each element, it performs the XOR operation with every other element in the list. This nested iteration results in a number of XOR operations proportional to n * n (approximately n * n/2, as each pair is effectively visited twice). Therefore, the time complexity is O(n²), indicating that the execution time grows quadratically with the input size n.
Space Complexity
O(1) – The provided plain English explanation outlines a brute-force approach that iterates through pairs of numbers and calculates XOR values. It does not mention any auxiliary data structures like lists or hash maps for storing intermediate results. Only a running total (which occupies constant space) and loop counters are implicitly used. Therefore, the space complexity remains constant regardless of the input size N (the number of elements in the list).
Optimal Solution
Approach
Calculating the XOR of all possible pairs directly is slow. The key idea is that we only care about how many times each number appears in the final XOR sum. If a number appears an even number of times, it cancels itself out in the XOR operation.
Here's how the algorithm would work step-by-step:
- Count how many times each number in the input appears.
- For each number, check if its count is odd or even.
- If a number appears an odd number of times, include that number in the final XOR sum.
- If a number appears an even number of times, it doesn't affect the final XOR sum (because XORing it an even number of times is the same as XORing it with zero).
- Calculate the XOR of all the numbers that appear an odd number of times. This is your final answer.
Code Implementation
def bitwise_xor_of_all_pairings(numbers):
number_counts = {}
for number in numbers:
if number in number_counts:
number_counts[number] += 1
else:
number_counts[number] = 1
xor_sum = 0
# Only numbers with odd counts contribute to the final XOR sum
for number, count in number_counts.items():
if count % 2 != 0:
# XOR in numbers that appear an odd amount of times.
xor_sum ^= number
return xor_sumBig(O) Analysis
Time Complexity
O(n) – The provided solution involves counting the occurrences of each number in the input array of size n. This typically requires iterating through the array once, which takes O(n) time. Next, the solution iterates through the unique numbers to determine their frequency parity (odd or even), which is bounded by n. Finally, it performs XOR operations on the numbers appearing an odd number of times, which again is bounded by n. Therefore, the overall time complexity is dominated by the initial counting and subsequent frequency analysis, making it O(n).
Space Complexity
O(N) – The algorithm uses a hash map to count the occurrences of each number in the input array. In the worst-case scenario, all N numbers in the input array are distinct, requiring the hash map to store N key-value pairs. Therefore, the space required by the hash map grows linearly with the input size N. Consequently, the auxiliary space complexity is O(N).
Edge Cases
| Case | How to Handle |
|---|---|
| Null or empty input array | Return 0 immediately as there are no pairings to XOR. |
| Array with only one element | Return 0 immediately since a pairing requires at least two elements. |
| Array with two identical elements | The XOR of the two elements is calculated correctly as 0, and the result is returned. |
| Large array exceeding memory constraints | The optimal solution uses the mathematical property that (a XOR b) XOR (a XOR c) XOR ... XOR (a XOR n) = a XOR a XOR ... XOR a XOR (b XOR c XOR ... XOR n), leading to a time complexity of O(n) and space complexity of O(1). |
| Array contains only zeros | The result will be 0 since 0 XOR 0 is always 0, handled correctly by the XOR operations. |
| Array containing very large numbers close to the maximum integer value | Ensure the integer type used to store the XOR result is large enough to avoid overflow by using long if necessary. |
| Array contains negative numbers (if language supports them) | The XOR operation works correctly with negative numbers based on their bit representation. |
| Array contains a mixture of positive, negative and zero values | The XOR operations handle different types of numbers correctly according to their bit representations. |
Null or empty input array
How to Handle:
Return 0 immediately as there are no pairings to XOR.
Array with only one element
How to Handle:
Return 0 immediately since a pairing requires at least two elements.
Array with two identical elements
How to Handle:
The XOR of the two elements is calculated correctly as 0, and the result is returned.
Large array exceeding memory constraints
How to Handle:
The optimal solution uses the mathematical property that (a XOR b) XOR (a XOR c) XOR ... XOR (a XOR n) = a XOR a XOR ... XOR a XOR (b XOR c XOR ... XOR n), leading to a time complexity of O(n) and space complexity of O(1).
Array contains only zeros
How to Handle:
The result will be 0 since 0 XOR 0 is always 0, handled correctly by the XOR operations.
Array containing very large numbers close to the maximum integer value
How to Handle:
Ensure the integer type used to store the XOR result is large enough to avoid overflow by using long if necessary.
Array contains negative numbers (if language supports them)
How to Handle:
The XOR operation works correctly with negative numbers based on their bit representation.
Array contains a mixture of positive, negative and zero values
How to Handle:
The XOR operations handle different types of numbers correctly according to their bit representations.
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