# Number of solutions of n = x + n ⊕ x

Given a number n, we have to find the number of possible values of X such that n = x + n ⊕ x. Here ⊕ represents XOR

Examples:

```Input : n = 3
Output : 4
The possible values of x are 0, 1, 2, and 3.

Input : n = 2
Output : 2
The possible values of x are 0 and 2.
```

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

Brute force approach: We can see that x is always equal to or less than n, so we can iterate over the range [0, n] and count the number of values that satisfy the required condition. The time complexity of this approach is O(n).

## C++

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the number of ` `// solutions of n = n xor x ` `int` `numberOfSolutions(``int` `n) ` `{ ` `    ``// Counter to store the number ` `    ``// of solutions found ` `    ``int` `c = 0; ` ` `  `    ``for` `(``int` `x = 0; x <= n; ++x) ` `        ``if` `(n == x + n ^ x) ` `            ``++c; ` ` `  `    ``return` `c; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 3; ` `    ``cout << numberOfSolutions(n); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of above approach ` `import` `java.util.*; ` `import` `java.lang.*; ` ` `  `class` `GFG ` `{ ` `// Function to find the number of ` `// solutions of n = n xor x ` `static` `int` `numberOfSolutions(``int` `n) ` `{ ` `    ``// Counter to store the number ` `    ``// of solutions found ` `    ``int` `c = ``0``; ` ` `  `    ``for` `(``int` `x = ``0``; x <= n; ++x) ` `        ``if` `(n == x + (n ^ x)) ` `            ``++c; ` ` `  `    ``return` `c; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `n = ``3``; ` `    ``System.out.print(numberOfSolutions(n)); ` `} ` `} ` ` `  `// This code is contributed  ` `// by Akanksha Rai(Abby_akku) `

## Python 3

 `# Python 3 implementation  ` `# of above approach ` ` `  `# Function to find the number of ` `# solutions of n = n xor x ` `def` `numberOfSolutions(n): ` ` `  `    ``# Counter to store the number ` `    ``# of solutions found ` `    ``c ``=` `0` ` `  `    ``for` `x ``in` `range``(n ``+` `1``): ` `        ``if` `(n ``=``=``( x ``+``( n ^ x))): ` `            ``c ``+``=` `1` ` `  `    ``return` `c ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``n ``=` `3` `    ``print``(numberOfSolutions(n)) ` ` `  `# This code is contributed  ` `# by ChitraNayal `

## C#

 `// C# implementation of above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `// Function to find the number of ` `// solutions of n = n xor x ` `static` `int` `numberOfSolutions(``int` `n) ` `{ ` `    ``// Counter to store the number ` `    ``// of solutions found ` `    ``int` `c = 0; ` ` `  `    ``for` `(``int` `x = 0; x <= n; ++x) ` `        ``if` `(n == x + (n ^ x)) ` `            ``++c; ` ` `  `    ``return` `c; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `n = 3; ` `    ``Console.Write(numberOfSolutions(n)); ` `} ` `} ` ` `  `// This code is contributed  ` `// by Akanksha Rai(Abby_akku) `

## PHP

 `

Output:

```4
```

Time complexity: O(n)

Efficient approach: We can solve this problem in a more efficient way if we consider n in its binary form. If a bit of n is set, i.e. 1, then we can deduce that there must be a corresponding set bit in either x or n ⊕ x (but not both). If the corresponding bit is set in x, then it is not set in n ⊕ x as 1 ⊕ 1 = 0. Otherwise the bit is set in n ⊕ x as 0 ⊕ 1 = 1. Therefore for every set bit in n, we can have either a set bit or an unset bit in x. However, we cannot have a set bit in x corresponding to an unset bit in n. By this logic, the number of solutions comes out to be 2 raised to the power of the number of set bits in n. The time complexity of this approach is O(log n).

## C++

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the number of ` `// solutions of n = n xor x ` `int` `numberOfSolutions(``int` `n) ` `{ ` `    ``// Number of set bits in n ` `    ``int` `c = 0; ` ` `  `    ``while` `(n) { ` `        ``c += n % 2; ` `        ``n /= 2; ` `    ``} ` ` `  `    ``// We can also write (1 << c) ` `    ``return` `pow``(2, c); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 3; ` `    ``cout << numberOfSolutions(n); ` `    ``return` `0; ` `} `

## Java

 `// Java  implementation of above approach ` `import` `java.io.*; ` ` `  `class` `GFG { ` `// Function to find the number of ` `// solutions of n = n xor x ` `static` `int` `numberOfSolutions(``int` `n) ` `{ ` `    ``// Number of set bits in n ` `    ``int` `c = ``0``; ` ` `  `    ``while` `(n>``0``) { ` `        ``c += n % ``2``; ` `        ``n /= ``2``; ` `    ``} ` ` `  `    ``// We can also write (1 << c) ` `    ``return` `(``int``)Math.pow(``2``, c); ` `} ` ` `  `// Driver code ` ` `  `    ``public` `static` `void` `main (String[] args) { ` `        ``int` `n = ``3``; ` `    ``System.out.println( numberOfSolutions(n)); ` `    ``} ` `} ` `//This code is contributed by anuj_67 `

## Python 3

 `# Python3 implementation of above approach  ` ` `  `# from math lib. import everything ` `from` `math ``import` `*` ` `  `# Function to find the number of  ` `# solutions of n = n xor x  ` `def` `numberOfSolutions(n) : ` ` `  `    ``# Number of set bits in n  ` `    ``c ``=` `0` ` `  `    ``while``(n) : ` `        ``c ``+``=` `n ``%` `2` `        ``n ``/``/``=` `2` ` `  `    ``# We can also write (1 << c)  ` `    ``return` `int``(``pow``(``2``, c)) ` ` `  `         `  `# Driver code      ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``n ``=` `3` `    ``print``(numberOfSolutions(n)) ` ` `  `# This code is contributed by ANKITRAI1 `

## C#

 `// C# implementation of above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `// Function to find the number of ` `// solutions of n = n xor x ` `static` `int` `numberOfSolutions(``int` `n) ` `{ ` `    ``// Number of set bits in n ` `    ``int` `c = 0; ` ` `  `    ``while` `(n > 0)  ` `    ``{ ` `        ``c += n % 2; ` `        ``n /= 2; ` `    ``} ` ` `  `    ``// We can also write (1 << c) ` `    ``return` `(``int``)Math.Pow(2, c); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main ()  ` `{ ` `    ``int` `n = 3; ` `    ``Console.WriteLine(numberOfSolutions(n)); ` `} ` `} ` ` `  `// This code is contributed by anuj_67 `

## PHP

 ` `

Output:

```4
```

Time complexity: O(log n)

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