Given a integer ‘x’, find the number of values of ‘a’ satisfying the following conditions:

- a XOR x > x
- 0 < a < x

Examples:

Input : x = 10 Output : 5 Explanation: For x = 10, following 5 values of 'a' satisfy the conditions: 1 XOR 10 = 11 4 XOR 10 = 14 5 XOR 10 = 15 6 XOR 10 = 12 7 XOR 10 = 13 Input : x = 2 Output : 1 Explanation: For x=2, we have just one value 1 XOR 2 = 3.

**Naive Approach**

A Simple approach is to check for all values of ‘a’ between 0 and ‘x’ and calculate its XOR with x

and check if the condition 1 satisfies.

## C++

// C++ program to find count of values // whose XOR with x is greater than x // and values are smaller than x #include<bits/stdc++.h> using namespace std; int countValues(int x) { int count = 0; for (int i=1; i < x; i++) if ((i ^ x) > x) count++; return count; } // Driver code int main() { int x = 10; cout << countValues(x); return 0; }

## Java

// Java program to find count of values // whose XOR with x is greater than x // and values are smaller than x public class XOR { static int countValues(int x) { int count = 0; for (int i=1; i < x; i++) if ((i ^ x) > x) count++; return count; } public static void main (String[] args) { int x = 10; System.out.println(countValues(x)); } } // This code is contributed by Saket Kumar

Output :

5

The time complexity of the above approach is O(x).

**Efficient Approach**

Efficient solution lies in binary representation of the number. We consider all 0’s in binary representation. For every 0 at i-th position, we can have 2^{i} numbers smaller than or equal to x with greater XOR.

## C++

// C++ program to find count of values // whose XOR with x is greater than x // and values are smaller than x #include<bits/stdc++.h> using namespace std; int countValues(int x) { // Initialize result int count = 0, n = 1; // Traversing through all bits of x while (x != 0) { // If current last bit of x is set // then increment count by n. Here // n is a power of 2 corresponding // to position of bit if (x%2 == 0) count += n; // Simultaneously calculate the 2^n n *= 2; // Replace x with x/2; x /= 2; } return count; } // Driver code int main() { int x = 10; cout << countValues(x); return 0; }

## Java

// Java program to find count of values // whose XOR with x is greater than x // and values are smaller than x public class XOR { static int countValues(int x) { // Initialize result int count = 0, n = 1; // Traversing through all bits of x while (x != 0) { // If current last bit of x is set // then increment count by n. Here // n is a power of 2 corresponding // to position of bit if (x%2 == 0) count += n; // Simultaneously calculate the 2^n n *= 2; // Replace x with x/2; x /= 2; } return count; } public static void main (String[] args) { int x = 10; System.out.println(countValues(x)); } } // This code is contributed by Saket Kumar

Output:

5

Time complexity of this solution is O(Log x)

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