Bitwise XOR of All Pairings
A medium-tier problem at 67% community acceptance, tagged with Array, Bit Manipulation, Brainteaser. Reported in interviews at Trilogy and 0 others.
Bitwise XOR of All Pairings is a medium-difficulty problem that looks like a brute-force pairing nightmare until you see the trick. You're given two arrays and need to XOR every element from the first array with every element from the second, then XOR all results together. The naive approach is O(n*m), but the actual solution hinges on a single bit-manipulation insight that collapses the problem entirely. It's been asked at Trilogy and appears in most platform rotation cycles. If you haven't drilled the XOR-count pattern, StealthCoder solves it in seconds during your live assessment.
Companies that ask "Bitwise XOR of All Pairings"
Bitwise XOR of All Pairings is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Built by an engineer at a top-10 tech company who can solve these problems cold but didn't want to trust himself in a 90-minute screen share.
Get StealthCoderThe key insight is that XOR has a cyclic property: if an element appears an even number of times in the full list of pairings, it cancels out to zero. Every element in the first array gets paired with every element in the second array exactly once. So element arr1[i] appears len(arr2) times in the full XOR operation, and arr2[j] appears len(arr1) times. If both lengths are even, all elements cancel and the answer is zero. If both are odd, you XOR all elements from both arrays. If one is odd and one is even, only the odd-length array contributes. The trap is coding a nested loop first and watching your solution time out. StealthCoder is the hedge if you blank on whether to count parity or actually compute pairings.
Pattern tags
You know the problem.
Make sure you actually pass it.
Bitwise XOR of All Pairings recycles across companies for a reason. It's medium-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Built by an engineer at a top-10 tech company who can solve these problems cold but didn't want to trust himself in a 90-minute screen share. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Bitwise XOR of All Pairings interview FAQ
Can I just XOR all pairs in a nested loop?+
Technically yes, but it's O(n*m) and will time out on large inputs. The problem expects you to recognize the parity pattern and skip the loop entirely. If both array lengths are even, return 0. If both are odd, XOR both arrays and return the result. That's O(n + m).
Why does XOR cancellation work here?+
XOR is self-inverse: a XOR a = 0. If an element appears an even number of times across all pairs, it cancels. Count how many times each element from arr1 and arr2 is actually XORed. That count is determined purely by array lengths, not values.
What if one array is empty?+
If either array is empty, there are no pairs to form, so the result is 0. But check the problem constraints. Most versions specify non-empty arrays, so this may not be a test case.
Is this still asked at FAANG?+
It's confirmed at Trilogy. Medium-difficulty bit-manipulation problems are standard at most large tech companies. If you see this in your assessment, it's likely in their rotation regularly.
How does this relate to other XOR problems?+
Like other XOR puzzles, it exploits the fact that XOR is associative and commutative, and pairs of identical values cancel. The novelty here is recognizing that the problem structure (all pairings) determines counts without enumeration. Common on arrays and bit-manipulation problem sets.
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