# 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: 9
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; ` `} `

Output:

```9
```

Time complexity: O(log(X))

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