# Count of pairs having bit size at most X and Bitwise OR equal to X

Given a number X, calculate number of possible pairs (a, b) such that bitwise or of a and b is equal to X and number of bits in both a and b is less than equal to number of bits in X.

Examples:

Input: X = 6
Output:
Explanation:
The possible pairs of (a, b) are (4, 6), (6, 4), (6, 6), (6, 2), (4, 2), (6, 0), (2, 6), (2, 4), (0, 6).

Input: X = 21
Output: 27
Explanation:
In total there are 27 pairs possible.

Approach: To solve the problem mentioned above follow the steps given below:

• Iterate through every bit of given number X.
• If the bit is 1 then from the truth table of Bitwise OR we know that there are 3 combinations possible for that given bit in number a and b that is (0, 1), (1, 0), (1, 1) that is 3 possible ways.
• If the bit is 0 then from the truth table of Bitwise OR we know that there is only 1 combination possible for that given bit in number a and b that is (0, 0).
• So our answer will be answer will be 3 ^ (number of on bits in X).

Below is the implementation of above approach:

 `// C++ implementation to Count number of ` `// possible pairs of (a, b) such that ` `// their Bitwise OR gives the value X ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to count the pairs ` `int` `count_pairs(``int` `x) ` `{ ` `    ``// Initializing answer with 1 ` `    ``int` `ans = 1; ` ` `  `    ``// Iterating through bits of x ` `    ``while` `(x > 0) { ` ` `  `        ``// check if bit is 1 ` `        ``if` `(x % 2 == 1) ` ` `  `            ``// multiplying ans by 3 ` `            ``// if bit is 1 ` `            ``ans = ans * 3; ` ` `  `        ``x = x / 2; ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `X = 6; ` ` `  `    ``cout << count_pairs(X) ` `         ``<< endl; ` ` `  `    ``return` `0; ` `} `

 `// Java implementation to count number of ` `// possible pairs of (a, b) such that ` `// their Bitwise OR gives the value X ` `class` `GFG{ ` ` `  `// Function to count the pairs ` `static` `int` `count_pairs(``int` `x) ` `{ ` `     `  `    ``// Initializing answer with 1 ` `    ``int` `ans = ``1``; ` ` `  `    ``// Iterating through bits of x ` `    ``while` `(x > ``0``) ` `    ``{ ` `         `  `        ``// Check if bit is 1 ` `        ``if` `(x % ``2` `== ``1``) ` ` `  `            ``// Multiplying ans by 3 ` `            ``// if bit is 1 ` `            ``ans = ans * ``3``; ` ` `  `        ``x = x / ``2``; ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `X = ``6``; ` ` `  `    ``System.out.print(count_pairs(X) + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by amal kumar choubey `

 `# Python3 implementation to count number of ` `# possible pairs of (a, b) such that ` `# their Bitwise OR gives the value X ` ` `  `# Function to count the pairs ` `def` `count_pairs(x): ` ` `  `    ``# Initializing answer with 1 ` `    ``ans ``=` `1``; ` ` `  `    ``# Iterating through bits of x ` `    ``while` `(x > ``0``): ` ` `  `        ``# Check if bit is 1 ` `        ``if` `(x ``%` `2` `=``=` `1``): ` ` `  `            ``# Multiplying ans by 3 ` `            ``# if bit is 1 ` `            ``ans ``=` `ans ``*` `3``; ` ` `  `        ``x ``=` `x ``/``/` `2``; ` `     `  `    ``return` `ans; ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `     `  `    ``X ``=` `6``; ` ` `  `    ``print``(count_pairs(X)); ` ` `  `# This code is contributed by amal kumar choubey  `

 `// C# implementation to count number of ` `// possible pairs of (a, b) such that ` `// their Bitwise OR gives the value X ` `using` `System; ` `class` `GFG{ ` ` `  `  ``// Function to count the pairs ` `  ``static` `int` `count_pairs(``int` `x)  ` `  ``{ ` ` `  `    ``// Initializing answer with 1 ` `    ``int` `ans = 1; ` ` `  `    ``// Iterating through bits of x ` `    ``while` `(x > 0)  ` `    ``{ ` ` `  `      ``// Check if bit is 1 ` `      ``if` `(x % 2 == 1) ` ` `  `        ``// Multiplying ans by 3 ` `        ``// if bit is 1 ` `        ``ans = ans * 3; ` ` `  `      ``x = x / 2; ` `    ``} ` `    ``return` `ans; ` `  ``} ` ` `  `  ``// Driver code ` `  ``public` `static` `void` `Main(String[] args)  ` `  ``{ ` `    ``int` `X = 6; ` ` `  `    ``Console.Write(count_pairs(X) + ``"\n"``); ` `  ``} ` `} ` ` `  `// This code is contributed by sapnasingh4991`

Output:
```9
```

Time complexity: O(log(X))

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