Given two integers **X** and **Y**, the task is to find the total number of ways to generate a pair of integers **A** and **B** such that Bitwise XOR and Bitwise AND between **A** and **B** is **X **and **Y** respectively

**Examples:**

Input:X = 2, Y = 5Output:2Explanation:

The two possible pairs are (5, 7) and (7, 5).Pair 1:(5, 7)

Bitwise AND = 5 & 7 = 2

Bitwise XOR = 5 ^ 7 = 5Pair 2:(7, 5)

Bitwise AND = 7 & 5 = 2

Bitwise XOR = 7 ^ 5 = 5

Input:X = 7, Y = 5Output:0

**Naive Approach:** The simplest approach to solve the problem is to choose the maximum among **X** and **Y **and set all its bits and then check all possible pairs from **0** to that maximum number, say **M**. If for any pair of **A** and **B**, **A & B** and **A⊕B **becomes equal to **X** and **Y** respectively, then increment the **count**. Print the final value of **count** after checking for all possible pairs.**Time Complexity:** O(M^{2})**Auxiliary Space:** O(1)

**Efficient Approach:** The idea is to generate all possible combinations of bits at each position. There are 4 possibilities for the **i ^{th}** bit in

**X**and

**Y**which are as follows:

- If X
_{i}= 0 and Y_{i}= 1, then A_{i}= 1 and B_{i}= 1. - If X
_{i}= 1 and Y_{i}= 1, then no possible bit assignment exist. - If X
_{i}= 0 and Y_{i}= 0 then A_{i}= 0 and B_{i}= 0. - If X
_{i}= 1 and Y_{i}= 0 then A_{i}= 0 and B_{i}= 1 or A_{i}= 1 and B_{i}= 0, where**A**,_{i}**B**,_{i}**X**, and_{i}**Y**represent_{i}**i**bit in each of them.^{th}

Follow the steps below to solve the problem:

- Initialize the counter as
**1**. - For the
**i**, if^{th}bit**X**and_{i}**Y**are equal to_{i}**1**, then print**0**. - If at
**i**,^{th}bit**X**is_{i}**1**and**Y**is_{i}**0**then multiply the counter by**2**as there are 2 options. - After that, divide
**X**and**Y**each time by**2**. - Repeat the above steps for every
**i**until both of them become^{th}bit**0**and print the value of the**counter**.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <iostream>` `using` `namespace` `std;` `// Function to return the count of` `// possible pairs of A and B whose` `// Bitwise XOR is X and Y respectively` `int` `countOfPairs(` `int` `x, ` `int` `y)` `{` ` ` `// Stores the count of pairs` ` ` `int` `counter = 1;` ` ` `// Iterate till any bit are set` ` ` `while` `(x || y) {` ` ` `// Extract i-th bit` ` ` `// of X and Y` ` ` `int` `bit1 = x % 2;` ` ` `int` `bit2 = y % 2;` ` ` `// Divide X and Y by 2` ` ` `x >>= 1;` ` ` `y >>= 1;` ` ` `// If Xi = 1 and Yi = 2,` ` ` `// multiply counter by 2` ` ` `if` `(bit1 == 1 and bit2 == 0) {` ` ` `// Increase required count` ` ` `counter *= 2;` ` ` `continue` `;` ` ` `}` ` ` `// If Xi =1 and Yi = 1` ` ` `if` `(bit1 & bit2) {` ` ` `// No answer exists` ` ` `counter = 0;` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `// Return the final count` ` ` `return` `counter;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given X and Y` ` ` `int` `X = 2, Y = 5;` ` ` `// Function Call` ` ` `cout << countOfPairs(X, Y);` ` ` `return` `0;` `}` |

## Java

`// Java program for` `// the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to return the count of` `// possible pairs of A and B whose` `// Bitwise XOR is X and Y respectively` `static` `int` `countOfPairs(` `int` `x, ` `int` `y)` `{` ` ` `// Stores the count of pairs` ` ` `int` `counter = ` `1` `;` ` ` `// Iterate till any bit are set` ` ` `while` `(x > ` `0` `|| y > ` `0` `)` ` ` `{` ` ` `// Extract i-th bit` ` ` `// of X and Y` ` ` `int` `bit1 = x % ` `2` `;` ` ` `int` `bit2 = y % ` `2` `;` ` ` `// Divide X and Y by 2` ` ` `x >>= ` `1` `;` ` ` `y >>= ` `1` `;` ` ` `// If Xi = 1 and Yi = 2,` ` ` `// multiply counter by 2` ` ` `if` `(bit1 == ` `1` `&& bit2 == ` `0` `)` ` ` `{` ` ` `// Increase required count` ` ` `counter *= ` `2` `;` ` ` `continue` `;` ` ` `}` ` ` `// If Xi =1 and Yi = 1` ` ` `if` `((bit1 & bit2) > ` `0` `)` ` ` `{` ` ` `// No answer exists` ` ` `counter = ` `0` `;` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `// Return the final count` ` ` `return` `counter;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `// Given X and Y` ` ` `int` `X = ` `2` `, Y = ` `5` `;` ` ` `// Function Call` ` ` `System.out.print(countOfPairs(X, Y));` `}` `}` `// This code is contributed by Princi Singh` |

## Python3

`# Python3 program for the above approach` `# Function to return the count of` `# possible pairs of A and B whose` `# Bitwise XOR is X and Y respectively` `def` `countOfPairs(x, y):` ` ` `# Stores the count of pairs` ` ` `counter ` `=` `1` ` ` `# Iterate till any bit are set` ` ` `while` `(x ` `or` `y):` ` ` `# Extract i-th bit` ` ` `# of X and Y` ` ` `bit1 ` `=` `x ` `%` `2` ` ` `bit2 ` `=` `y ` `%` `2` ` ` `# Divide X and Y by 2` ` ` `x >>` `=` `1` ` ` `y >>` `=` `1` ` ` `# If Xi = 1 and Yi = 2,` ` ` `# multiply counter by 2` ` ` `if` `(bit1 ` `=` `=` `1` `and` `bit2 ` `=` `=` `0` `):` ` ` ` ` `# Increase required count` ` ` `counter ` `*` `=` `2` ` ` `continue` ` ` `# If Xi =1 and Yi = 1` ` ` `if` `(bit1 & bit2):` ` ` `# No answer exists` ` ` `counter ` `=` `0` ` ` `break` ` ` `# Return the final count` ` ` `return` `counter` `# Driver Code` `# Given X and Y` `X ` `=` `2` `Y ` `=` `5` `# Function call` `print` `(countOfPairs(X, Y))` `# This code is contributed by Shivam Singh` |

## C#

`// C# program for` `// the above approach` `using` `System;` `class` `GFG{` `// Function to return the count of` `// possible pairs of A and B whose` `// Bitwise XOR is X and Y respectively` `static` `int` `countOfPairs(` `int` `x, ` `int` `y)` `{` ` ` `// Stores the count of pairs` ` ` `int` `counter = 1;` ` ` `// Iterate till any bit are set` ` ` `while` `(x > 0 || y > 0)` ` ` `{` ` ` `// Extract i-th bit` ` ` `// of X and Y` ` ` `int` `bit1 = x % 2;` ` ` `int` `bit2 = y % 2;` ` ` `// Divide X and Y by 2` ` ` `x >>= 1;` ` ` `y >>= 1;` ` ` `// If Xi = 1 and Yi = 2,` ` ` `// multiply counter by 2` ` ` `if` `(bit1 == 1 && bit2 == 0)` ` ` `{` ` ` `// Increase required count` ` ` `counter *= 2;` ` ` `continue` `;` ` ` `}` ` ` `// If Xi =1 and Yi = 1` ` ` `if` `((bit1 & bit2) > 0)` ` ` `{` ` ` `// No answer exists` ` ` `counter = 0;` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `// Return the readonly count` ` ` `return` `counter;` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` `// Given X and Y` ` ` `int` `X = 2, Y = 5;` ` ` `// Function Call` ` ` `Console.Write(countOfPairs(X, Y));` `}` `}` ` ` `// This code is contributed by Rajput-Ji` |

## Javascript

`<script>` `// JavaScript program for` `// the above approach` `// Function to return the count of` `// possible pairs of A and B whose` `// Bitwise XOR is X and Y respectively` `function` `countOfPairs(x, y)` `{` ` ` `// Stores the count of pairs` ` ` `let counter = 1;` ` ` ` ` `// Iterate till any bit are set` ` ` `while` `(x > 0 || y > 0)` ` ` `{` ` ` `// Extract i-th bit` ` ` `// of X and Y` ` ` `let bit1 = x % 2;` ` ` `let bit2 = y % 2;` ` ` ` ` `// Divide X and Y by 2` ` ` `x >>= 1;` ` ` `y >>= 1;` ` ` ` ` `// If Xi = 1 and Yi = 2,` ` ` `// multiply counter by 2` ` ` `if` `(bit1 == 1 && bit2 == 0)` ` ` `{` ` ` `// Increase required count` ` ` `counter *= 2;` ` ` `continue` `;` ` ` `}` ` ` ` ` `// If Xi =1 and Yi = 1` ` ` `if` `((bit1 & bit2) > 0)` ` ` `{` ` ` `// No answer exists` ` ` `counter = 0;` ` ` `break` `;` ` ` `}` ` ` `}` ` ` ` ` `// Return the final count` ` ` `return` `counter;` `}` `// Driver code` ` ` `// Given X and Y` ` ` `let X = 2, Y = 5;` ` ` ` ` `// Function Call` ` ` `document.write(countOfPairs(X, Y));` ` ` `</script>` |

**Output:**

2

**Time Complexity:** O(log M)**Auxiliary Space:** O(1)

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