# Equal Sum and XOR of three Numbers

Given an integer N. The task is to count the numbers of pairs of integers A and B such that A + B + N = A ^ B ^ N and A and B are less than N.

Examples:

Input: N = 2
Output: 3
Explanation:-
For N = 2
2 XOR 0 XOR 0 = 2+0+0
2 XOR 0 XOR 1 = 2+0+1
2 XOR 0 XOR 2 != 2+0+2
2 XOR 1 XOR 0 = 2+1+0
2 XOR 1 XOR 1 != 2+1+1
2 XOR 1 XOR 2 != 2+1+2
2 XOR 2 XOR 0 != 2+2+0
2 XOR 2 XOR 1 != 2+2+1
2 XOR 2 XOR 2 != 2+2+2

So (0, 0), (0, 1) and (1, 0) are the required pairs. So the output is 3.

Input: N = 4
Output: 9

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

Approach :
To make the sum of three numbers equal to the xor of three number with one of the number given we can do following:-

1. Represent the fixed number in binary form.
2. Traverse the binary expansion of the fixed number.
• If you find a 1 there is only one condition i.e. you take the other two number’s binary bits as 0 and 0.
• If you find a 0 there will be three conditions i.e. either you can have binary bits as (0, 0), (1, 0)
or (0, 1).
3. The following above triplets of bits will never go for a carry so they are valid.
4. So the answer will be 3^(number of zeros in binary representation).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Defining ull to unsigned long long int ` `typedef` `unsigned ``long` `long` `int` `ull; ` ` `  `// Function to calculate power of 3 ` `ull calculate(``int` `bit_cnt) ` `{ ` `    ``ull res = 1; ` `    ``while` `(bit_cnt--) { ` `        ``res = res * 3; ` `    ``} ` ` `  `    ``return` `res; ` `} ` ` `  `// Function to return the count of the ` `// unset bit ( zeros ) ` `int` `unset_bit_count(ull n) ` `{ ` `    ``int` `count = 0; ` `    ``while` `(n) { ` ` `  `        ``// Check the bit is 0 or not ` `        ``if` `((n & 1) == 0) ` `            ``count++; ` `        ``// Right shifting ( dividing by 2 ) ` `        ``n = n >> 1; ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``ull n; ` `    ``n = 2; ` ` `  `    ``int` `count = unset_bit_count(n); ` ` `  `    ``ull ans = calculate(count); ` ` `  `    ``cout << ans << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to calculate power of 3 ` `static` `long` `calculate(``int` `bit_cnt) ` `{ ` `    ``long` `res = ``1``; ` `    ``while` `(bit_cnt-- > ``0``)  ` `    ``{ ` `        ``res = res * ``3``; ` `    ``} ` ` `  `    ``return` `res; ` `} ` ` `  `// Function to return the count of the ` `// unset bit ( zeros ) ` `static` `int` `unset_bit_count(``long` `n) ` `{ ` `    ``int` `count = ``0``; ` `    ``while` `(n > ``0``)  ` `    ``{ ` ` `  `        ``// Check the bit is 0 or not ` `        ``if` `((n & ``1``) == ``0``) ` `            ``count++; ` `             `  `        ``// Right shifting ( dividing by 2 ) ` `        ``n = n >> ``1``; ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``long` `n; ` `    ``n = ``2``; ` ` `  `    ``int` `count = unset_bit_count(n); ` ` `  `    ``long` `ans = calculate(count); ` ` `  `    ``System.out.println(ans); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992  `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to calculate power of 3 ` `def` `calculate(bit_cnt): ` ` `  `    ``res ``=` `1``; ` `    ``while` `(bit_cnt > ``0``):  ` `        ``bit_cnt ``-``=` `1``; ` `        ``res ``=` `res ``*` `3``; ` `    ``return` `res; ` ` `  `# Function to return the count of the ` `# unset bit ( zeros ) ` `def` `unset_bit_count(n): ` ` `  `    ``count ``=` `0``; ` `    ``while` `(n > ``0``): ` `         `  `        ``# Check the bit is 0 or not ` `        ``if` `((n & ``1``) ``=``=` `0``): ` `            ``count ``+``=` `1``; ` `             `  `        ``# Right shifting ( dividing by 2 ) ` `        ``n ``=` `n >> ``1``; ` `     `  `    ``return` `count; ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` ` `  `    ``n ``=` `2``; ` ` `  `    ``count ``=` `unset_bit_count(n); ` ` `  `    ``ans ``=` `calculate(count); ` ` `  `    ``print``(ans); ` ` `  `# This code contributed by Rajput-Ji `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to calculate power of 3 ` `static` `long` `calculate(``int` `bit_cnt) ` `{ ` `    ``long` `res = 1; ` `    ``while` `(bit_cnt-- > 0)  ` `    ``{ ` `        ``res = res * 3; ` `    ``} ` ` `  `    ``return` `res; ` `} ` ` `  `// Function to return the count of the ` `// unset bit (zeros) ` `static` `int` `unset_bit_count(``long` `n) ` `{ ` `    ``int` `count = 0; ` `    ``while` `(n > 0)  ` `    ``{ ` ` `  `        ``// Check the bit is 0 or not ` `        ``if` `((n & 1) == 0) ` `            ``count++; ` `             `  `        ``// Right shifting ( dividing by 2 ) ` `        ``n = n >> 1; ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``long` `n; ` `    ``n = 2; ` ` `  `    ``int` `count = unset_bit_count(n); ` ` `  `    ``long` `ans = calculate(count); ` ` `  `    ``Console.WriteLine(ans); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```3
```

Time Complexity: O(Number of unset_bits) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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