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

• Last Updated : 22 Mar, 2021

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.```

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)`

## Python3

 `# 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

 `

## Javascript

 ``
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`

## Python3

 `# 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

 ``

## Javascript

 ``
Output:
`4`

Time complexity: O(log n)

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