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# Count numbers whose difference with N is equal to XOR with N

• Difficulty Level : Easy
• Last Updated : 28 Apr, 2021

Given a number N. The task is to count all possible values of x such that n x is equal to (N-x), where denotes bitwise XOR operation.
Examples:

```Input: N = 3
Output: 4
The all possible values of x are respectively 0, 1, 2, 3.

Input: N = 6
Output: 4
The all possible values of x are respectively 0, 2, 4, 6.```

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Approach: The XOR value of two bits will be 1 if both bits have opposite sign, and 0 when both bits are same. So on the basis of the property of XOR, we can say that n x is always greater than or equal to n-x. The only condition when its value is equal with n-x is bits of x form a subset of bits of n. Because if in the i’th position both x and n has set bits then after xor the value will decrease, and the decreased value will be , where i is 0-based position.
So the answer is the total count of subsets of bits of number n is , where k is the count of set bits in n.
Below is the implementation of above approach:

## C++

 `#include ``using` `namespace` `std;` `// function to Count all values of x``void` `count_values(``int` `n)``{``    ``// Count set bits in n``    ``// by using stl function``    ``int` `set_bits = __builtin_popcount(n);` `    ``// count all subset of set bits``    ``cout << ``pow``(2, set_bits) << ``"\n"``;``}` `// Driver code``int` `main()``{` `    ``int` `n = 27;``    ``count_values(n);` `    ``return` `0;``}`

## Java

 `import` `java.util.*;` `class` `Solution``{``//count number of set bits``static` `int` `__builtin_popcount(``int` `n)``{``    ``//count variable``    ``int` `count=``0``;``    ` `    ``while``(n>``0``)``    ``{``        ``//if the bit is 1``        ``if``(n%``2``==``1``)``        ``count++;``        ` `        ``n=n/``2``;``    ``}``    ``return` `count;``}``    ` `// function to Count all values of x``static` `void` `count_values(``int` `n)``{``    ``// Count set bits in n``    ``// by using stl function``    ``int` `set_bits = __builtin_popcount(n);``  ` `    ``// count all subset of set bits``    ``System.out.println((``int``)Math.pow(``2``, set_bits));``}``  ` `// Driver code``public` `static` `void` `main(String args[])``{``  ` `    ``int` `n = ``27``;``    ``count_values(n);``  `  `}``}` `// This code is contributed``// by Arnab Kundu`

## Python 3

 `# Python3 program to implement``# above approach` `# from math import pow method``from` `math ``import` `pow` `# count number of set bits``def` `__builtin_popcount(n) :` `    ``# count variable``    ``count ``=` `0` `    ``while` `n > ``0` `:` `        ``# if the bit is 1``        ``if` `n ``%` `2` `=``=` `1` `:``            ``count ``+``=` `1` `        ``n ``=` `n``/``/``2``        ` `    ``return` `count`  `# function to Count all values of x``def` `count_values(n) :` `    ``set_bits ``=` `__builtin_popcount(n)` `    ``# count all subset of set bits``    ``print``(``int``(``pow``(``2``, set_bits)))`  `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``n ``=` `27``    ``count_values(n)` `# This code is contributed by``# ANKITRAI1`

## C#

 `using` `System;``class` `GFG``{``// count number of set bits``static` `int` `__builtin_popcount(``int` `n)``{``    ``// count variable``    ``int` `count = 0;``    ` `    ``while``(n > 0)``    ``{``        ``//if the bit is 1``        ``if``(n % 2 == 1)``        ``count++;``        ` `        ``n = n / 2;``    ``}``    ``return` `count;``}``    ` `// function to Count all values of x``static` `void` `count_values(``int` `n)``{``    ``// Count set bits in n``    ``// by using stl function``    ``int` `set_bits = __builtin_popcount(n);``    ` `    ``// count all subset of set bits``    ``Console.Write((``int``)Math.Pow(2, set_bits));``}``    ` `// Driver code``public` `static` `void` `Main()``{``    ``int` `n = 27;``    ``count_values(n);``}``}` `// This code is contributed by Smitha`

## PHP

 ` 0)``    ``{``        ``//if the bit is 1``        ``if``(``\$n` `% 2 == 1)``            ``\$count``++;``        ` `        ``\$n` `= ``\$n` `/ 2;``    ``}``    ``return` `\$count``;``}``    ` `// function to Count all values of x``function` `count_values(``\$n``)``{``    ``// Count set bits in n``    ``// by using stl function``    ``\$set_bits` `= __builtin_popcount(``\$n``);` `    ``// count all subset of set bits``    ``echo` `(int)pow(2, ``\$set_bits``);``}` `// Driver code``\$n` `= 27;``count_values(``\$n``);` `// This code is contributed``// by Akanksha Rai(Abby_akku)``?>`

## Javascript

 ``
Output:
`16`

Time Complexity: O(k), where k is number of set bits in N.

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