it is a difficult to say why you have wrong answer ,but the main idea to solve this problem is different. You must precalculate this numbers . The size of this numbers is not big. Maximum 31*31. You can add this numbers to vector and then sort it. After that ,with binary search you must to take three numbers and then print minimum difference.

Your code doesn't work for the these numbers 15 63 127 255 511 have you got the pattern? These are one less than the power of two and greater than 7. Your code returns 1 but the actual answer is 2.

Bro, just an advice. Don't go with so much complex logics. It never helps but just increases chances of getting WA as evident in your case. Even I suffered a lot because of the same habit. So try to think easy.

Coming to the code, I'll think on this and let you know if I find the mistake.

Till then. What I thought of the solution to the problem is that till 10^9, there will be at most 450 such numbers which are the sum of two numbers (different powers of 2). And note the optimal number can even exceed 10^9. So for safety take till 2*10^9. And then apply binary search to find the closest such number or the optimal number and accordingly calculate the answer.

There is really no need for such complex code or binary search. You can just iterate through the values of 2^x + 2^y for (x != y), then you simply take the minimum difference between n and the computed sum. It should run in O(30*30).