Number of Bit Changes to Make Two Integers Equal
A easy-tier problem at 63% community acceptance, tagged with Bit Manipulation. Reported in interviews at ThoughtWorks and 0 others.
You need to count how many bits differ between two integers. ThoughtWorks asks this one. It's classified easy, but candidates still blank on it during live assessments because the obvious loop-based approach feels too slow, even though it works fine. The real trick is recognizing that XOR immediately exposes every bit that differs, then you just count the 1s. If you hit this problem in your OA and your brain locks up on the counting step, StealthCoder surfaces a working solution in seconds, invisible to the proctor.
Companies that ask "Number of Bit Changes to Make Two Integers Equal"
Number of Bit Changes to Make Two Integers Equal 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. Made by an Amazon engineer who watched the leaked-problem repo become an industry secret. He decided you should have it too.
Get StealthCoderThe pattern: XOR two numbers and count set bits. Most candidates instinctively loop and compare bit-by-bit using masks, which is correct but feels clunky. The faster mental move is to realize XOR of two numbers produces 1 exactly where bits differ. From there, counting 1s in a binary representation is the standard bit-counting problem. Common pitfall is overthinking the comparison logic instead of just running XOR first. Another miss: using slow bit-by-bit loops when Brian Kernighan's trick (n & (n-1) to knock off each set bit) or built-in popcount functions exist. The assessment accepts either approach, but recognizing XOR as the immediate first step is what separates "got it" from "let me think about this." StealthCoder handles both the pattern recognition and the counting implementation so you move on.
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Number of Bit Changes to Make Two Integers Equal recycles across companies for a reason. It's easy-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. Made by an Amazon engineer who watched the leaked-problem repo become an industry secret. He decided you should have it too. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Number of Bit Changes to Make Two Integers Equal interview FAQ
Is this really an easy problem or does it trick you on the OA?+
Genuinely easy once you see XOR. The acceptance rate at 62.6% reflects that many candidates second-guess the approach rather than the difficulty itself. On a live OA, the gap between 'I think I need to loop' and 'XOR then popcount' costs time you don't have.
What's the actual trick to this problem?+
XOR the two numbers immediately. The result has 1 in every position where the bits differ. Then count the 1s using bit operations or a built-in function. That's it. No graph, no DP, no greedy. Pure bit math.
Do I need to handle negative integers?+
The problem doesn't specify constraints in the input data. Assume 32-bit or 64-bit integers depending on the platform. XOR works the same way for negatives in two's complement, so the algorithm doesn't change. Just watch for potential overflow if you're writing in a strict-typing language.
Is counting set bits something I should memorize?+
Yes. Brian Kernighan's trick (n & (n-1) removes one set bit per iteration) is fast and elegant. Python and Java have built-ins (bin().count('1'), Integer.bitCount()). Know at least one method well enough to code it under pressure.
Does ThoughtWorks ask this in a specific context or loop in the interview?+
Input data shows one report from ThoughtWorks. Without more reports, it's unclear if this is a screening filter or part of a deeper round. Treat it as the tier-1 problem to solve clean and fast, then move to harder bit manipulation or system design depending on the flow.
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