# Count smaller values whose XOR with x is greater than x

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

1. a XOR x > x
2. 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.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 2i 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)

This article is contributed by DANISH KALEEM. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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