# Count of values of x <= n for which (n XOR x) = (n – x)

Given an integer n, the task is to find the number of possible values of 0 ≤ x ≤ n which satisfy n XOR x = n – x.

Examples:

Input: n = 5
Output: 4
Following values of x satisfy the equation
5 XOR 0 = 5 – 0 = 5
5 XOR 1 = 5 – 1 = 4
5 XOR 4 = 5 – 4 = 1
5 XOR 5 = 5 – 5 = 0

Input: n = 2
Output: 2

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

Naive approach: The easy approach is to check for all values from 0 to n (both inclusive) and finding whether they satisfy the equation. The below code implements this approach:

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of  ` `// valid values of x ` `static` `int` `countX(``int` `n) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``for` `(``int` `i = 0; i <= n; i++) ` `    ``{ ` ` `  `        ``// If n - x = n XOR x ` `        ``if` `(n - i == (n ^ i)) ` `                ``count++; ` `    ``} ` ` `  `        ``// Return the required count; ` `        ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 5; ` `    ``int` `answer = countX(n); ` `    ``cout << answer; ` `} ` ` `  `// This code is contributed by  ` `// Shivi_Aggarwal `

## Java

 `// Java implementation of the approach ` `public` `class` `GFG { ` ` `  `    ``// Function to return the count of  ` `    ``// valid values of x ` `    ``static` `int` `countX(``int` `n) ` `    ``{ ` `        ``int` `count = ``0``; ` ` `  `        ``for` `(``int` `i = ``0``; i <= n; i++) { ` ` `  `            ``// If n - x = n XOR x ` `            ``if` `(n - i == (n ^ i)) ` `                ``count++; ` `        ``} ` ` `  `        ``// Return the required count; ` `        ``return` `count; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n = ``5``; ` `        ``int` `answer = countX(n); ` `        ``System.out.println(answer); ` `    ``} ` `} `

## Python3

 `# Python3 implementation of the approach ` `import` `math as mt ` ` `  `# Function to return the count of ` `# valid values of x ` `def` `countX(n): ` `    ``count ``=` `0` ` `  `    ``for` `i ``in` `range``(n ``+` `1``): ` ` `  `        ``if` `n ``-` `i ``=``=` `(n ^ i): ` `            ``count ``+``=` `1` ` `  `    ``return` `count ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n ``=` `5` `    ``answer ``=` `countX(n) ` `    ``print``(answer) ` ` `  `# This code is contributed by  ` `# Mohit kumar 29 `

## C#

 `// C# implementation of the above approach  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` ` `  `    ``// Function to return the count of  ` `    ``// valid values of x  ` `    ``static` `int` `countX(``int` `n)  ` `    ``{  ` `        ``int` `count = 0;  ` ` `  `        ``for` `(``int` `i = 0; i <= n; i++) ` `        ``{  ` ` `  `            ``// If n - x = n XOR x  ` `            ``if` `(n - i == (n ^ i))  ` `                ``count++;  ` `        ``}  ` ` `  `        ``// Return the required count;  ` `        ``return` `count;  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``int` `n = 5;  ` `        ``int` `answer = countX(n);  ` `        ``Console.WriteLine(answer);  ` `    ``}  ` `}  ` ` `  `// This code is contributed by Ryuga `

## PHP

 ` `

Output:

```4
```

Time complexity: O(N)

Efficient Approach: Convert n to its binary representation. Now, for every 1 in the binary string whether we subtract 1 or 0 from it, it will be equivalent to XOR of 1 with 0 or 1 i.e.
(1 – 1) = (1 XOR 1) = 0
(1 – 0) = (1 XOR 0) = 1
But 0 doesn’t satisfy this condition. So, we only need to consider all the ones in the binary representation of n. Now, for every 1 there are two possibilities, either 0 or 1. Thus if we have m number of 1’s in n then our solution would be 2m.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of ` `// valid values of x ` `int` `countX(``int` `n) ` `{ ` `    ``// Convert n into binary String ` `    ``string binary = bitset<8>(n).to_string(); ` `     `  `    ``// To store the count of 1s ` `    ``int` `count = 0; ` `    ``for` `(``int` `i = 0; i < binary.length(); i++)  ` `    ``{ ` `        ``// If current bit is 1 ` `        ``if` `(binary.at(i) == ``'1'``) ` `            ``count++; ` `    ``} ` `     `  `    ``// Calculating answer ` `    ``int` `answer = (``int``)``pow``(2, count); ` `    ``return` `answer; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 5; ` `    ``int` `answer = countX(n); ` `    ``cout << (answer); ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## Java

 `// Java implementation of the approach ` `public` `class` `GFG { ` ` `  `    ``// Function to return the count of ` `    ``// valid values of x ` `    ``static` `int` `countX(``int` `n) ` `    ``{ ` `        ``// Convert n into binary String ` `        ``String binary = Integer.toBinaryString(n); ` ` `  `        ``// To store the count of 1s ` `        ``int` `count = ``0``; ` ` `  `        ``for` `(``int` `i = ``0``; i < binary.length(); i++) { ` ` `  `            ``// If current bit is 1 ` `            ``if` `(binary.charAt(i) == ``'1'``) ` `                ``count++; ` `        ``} ` ` `  `        ``// Calculating answer ` `        ``int` `answer = (``int``)Math.pow(``2``, count); ` `        ``return` `answer; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n = ``5``; ` `        ``int` `answer = countX(n); ` `        ``System.out.println(answer); ` `    ``} ` `} `

## Python 3

 `# Python3 implementation of the approach ` ` `  `# Function to return the count of ` `# valid values of x ` `def` `countX(n): ` ` `  `    ``# Convert n into binary String ` `    ``binary ``=` `"{0:b}"``.``format``(n) ` ` `  `    ``# To store the count of 1s ` `    ``count ``=` `0` ` `  `    ``for` `i ``in` `range``(``len``(binary)): ` ` `  `        ``# If current bit is 1 ` `        ``if` `(binary[i] ``=``=` `'1'``): ` `            ``count ``+``=` `1` ` `  `    ``# Calculating answer ` `    ``answer ``=` `int``(``pow``(``2``, count)) ` `    ``return` `answer ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` `     `  `    ``n ``=` `5` `    ``answer ``=` `countX(n) ` `    ``print``(answer) ` ` `  `# This code is contributed by ita_c `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to return the count of ` `// valid values of x ` `static` `int` `countX(``int` `n) ` `{ ` `    ``// Convert n into binary String ` `    ``string` `binary = Convert.ToString(n, 2); ` ` `  `    ``// To store the count of 1s ` `    ``int` `count = 0; ` ` `  `    ``for` `(``int` `i = 0; i < binary.Length; i++)  ` `    ``{ ` ` `  `        ``// If current bit is 1 ` `        ``if` `(binary[i] == ``'1'``) ` `            ``count++; ` `    ``} ` ` `  `    ``// Calculating answer ` `    ``int` `answer = (``int``)Math.Pow(2, count); ` `    ``return` `answer; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `n = 5; ` `    ``int` `answer = countX(n); ` `    ``Console.WriteLine(answer); ` `} ` `} ` ` `  `// This code is contributed ` `// by Akanksha Rai `

Output:

```4
```

Time complexity:

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