#PROBLEM LINK:

**Author:** Ayush Nagal

#DIFFICULTY:

MEDIUM

#PREREQUISITES:

Bit Manipulation

#PROBLEM:

Given a long integer x, we have to find the number of values of a such that 0<a<x and a⊕x >x.

#EXPLANATION:

Let’s take an example, let the number x be 111010111 . Let’s start traversing the bit string from right to left, i.e., from lsb to msb. We can’t change the first, second or third bits as this would produce a number less than x. When we come to the 4th bit which is 0, we flip it to 1 (using XOR) and then we have 2^n choices to change the bits after that where n is the number of bits after the 4th bit. Then we move on to the next bit which is 1. Similarly, we continue traversing and flip 0 every time we encounter it and calculate the possibilites (2^n) and keep adding them. Consider the following snippet:

```
while (x > 0) {
if ((x & 1) == 0)
s += pow(2, i);
i++;
x = x >> 1;
}
```

Where x is the given integer and i is used for number of bits.

#AUTHOR’S SOLUTION:

Author’s solution can be found here.