# 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 = 6Output:9Explanation:

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 = 21Output:27Explanation:

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++

`// C++ implementation to Count number of` `// possible pairs of (a, b) such that` `// their Bitwise OR gives the value X` `#include <iostream>` `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

`// 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

`# 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#

`// 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` |

## Javascript

`<script>` `// javascript implementation to count number of` `// possible pairs of (a, b) such that` `// their Bitwise OR gives the value X ` `// Function to count the pairs` ` ` `function` `count_pairs(x) {` ` ` `// Initializing answer with 1` ` ` `var` `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 = parseInt(x / 2);` ` ` `}` ` ` `return` `ans;` ` ` `}` ` ` `// Driver code` ` ` ` ` `var` `X = 6;` ` ` `document.write(count_pairs(X) + ` `"\n"` `);` `// This code contributed by Rajput-Ji` `</script>` |

**Output:**

9

**Time complexity:** O(log(X))