# Find a pair of numbers with set bit count as at most that of N and whose Bitwise XOR is N

Given a positive integer** N**, the task is to find the pair of integers **(X, Y)** such that the Bitwise XOR of **X** and **Y** is **N** and **X * Y** is maximum where the count of bits in **X** and **Y** is less than or equal to **N**.

**Examples:**

Input:N = 10Output:13 7Explanation:The case where X = 13 and Y = 7 is the most optimal choice as 13 xor 7 = 10 and 13 * 7 = 91 which is maximum possible.

Input:N = 45Output:50 31

**Approach:** The given problem can be solved using the following observations:

- If the
**i**bit of^{th}**N**is**0**, then the**i**bit of both^{th}**X**and**Y**must be either**0**or**1**. In order to maximize the product, set such bits as**1**. - If the
**i**bit of^{th}**N**is**1**, then one of the**i**bits of^{th }**X**or**Y**must be**1**and the other must be**0**. Since**N**must have a constant number of set bits, therefore the sum of X and Y must be constant. - If the sum of two numbers is constant, their product is maximum when the difference between the two numbers is minimized.

According to the above observations, initialize two integers **X** and **Y** as 0. In order to minimize the difference between **X** and **Y**, **X **must be equal to the **MSB _{N}** and

**Y**must be equal to

**N – MSB**where

_{N}**MSB**represents the Most Significant Bit. Also, for all the

**0**bits in

**N**, set the respective bits in both

**X**and

**Y**as

**1**.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the pair (X, Y) such` `// that X xor Y = N and the count of set` `// bits in X and Y is less than count of` `// set bit in N` `void` `maximizeProduct(` `int` `N)` `{` ` ` `// Stores MSB (Most Significant Bit)` ` ` `int` `MSB = (` `int` `)log2(N);` ` ` `// Stores the value of X` ` ` `int` `X = 1 << MSB;` ` ` `// Stores the value of Y` ` ` `int` `Y = N - (1 << MSB);` ` ` `// Traversing over all bits of N` ` ` `for` `(` `int` `i = 0; i < MSB; i++) {` ` ` `// If ith bit of N is 0` ` ` `if` `(!(N & (1 << i))) {` ` ` `// Set ith bit of X to 1` ` ` `X += 1 << i;` ` ` `// Set ith bit of Y to 1` ` ` `Y += 1 << i;` ` ` `}` ` ` `}` ` ` `// Print Answer` ` ` `cout << X << ` `" "` `<< Y;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 45;` ` ` `maximizeProduct(N);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.io.*;` `class` `GFG` `{` `// Function to find the pair (X, Y) such` `// that X xor Y = N and the count of set` `// bits in X and Y is less than count of` `// set bit in N` `static` `void` `maximizeProduct(` `int` `N)` `{` ` ` `// Stores MSB (Most Significant Bit)` ` ` `int` `MSB = (` `int` `)(Math.log(N) / Math.log(` `2` `));` ` ` ` ` `// Stores the value of X` ` ` `int` `X = ` `1` `<< MSB;` ` ` ` ` `// Stores the value of Y` ` ` `int` `Y = N - (` `1` `<< MSB);` ` ` ` ` `// Traversing over all bits of N` ` ` `for` `(` `int` `i = ` `0` `; i < MSB; i++) {` ` ` ` ` `// If ith bit of N is 0` ` ` `if` `((N & (` `1` `<< i))==` `0` `) {` ` ` ` ` `// Set ith bit of X to 1` ` ` `X += ` `1` `<< i;` ` ` ` ` `// Set ith bit of Y to 1` ` ` `Y += ` `1` `<< i;` ` ` `}` ` ` `}` ` ` ` ` `// Print Answer` ` ` `System.out.println(X+` `" "` `+Y);` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `45` `;` ` ` `maximizeProduct(N);` `}` `}` `// This code is contributed by dwivediyash` |

## Python3

`# python 3 program for the above approach` `import` `math` `# Function to find the pair (X, Y) such` `# that X xor Y = N and the count of set` `# bits in X and Y is less than count of` `# set bit in N` `def` `maximizeProduct(N):` ` ` `# Stores MSB (Most Significant Bit)` ` ` `MSB ` `=` `(` `int` `)(math.log2(N))` ` ` `# Stores the value of X` ` ` `X ` `=` `1` `<< MSB` ` ` `# / Stores the value of Y` ` ` `Y ` `=` `N ` `-` `(` `1` `<< MSB)` ` ` `# Traversing over all bits of N` ` ` `for` `i ` `in` `range` `(MSB):` ` ` `# If ith bit of N is 0` ` ` `if` `(` `not` `(N & (` `1` `<< i))):` ` ` `# / Set ith bit of X to 1` ` ` `X ` `+` `=` `1` `<< i` ` ` `# Set ith bit of Y to 1` ` ` `Y ` `+` `=` `1` `<< i` ` ` `# Print Answer` ` ` `print` `(X, Y)` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `N ` `=` `45` ` ` `maximizeProduct(N)` ` ` `# This code is contributed by ukasp.` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG {` `// Function to find the pair (X, Y) such` `// that X xor Y = N and the count of set` `// bits in X and Y is less than count of` `// set bit in N` `static` `void` `maximizeProduct(` `int` `N)` `{` ` ` ` ` `// Stores MSB (Most Significant Bit)` ` ` `int` `MSB = (` `int` `)(Math.Log(N) / Math.Log(2));` ` ` ` ` `// Stores the value of X` ` ` `int` `X = 1 << MSB;` ` ` ` ` `// Stores the value of Y` ` ` `int` `Y = N - (1 << MSB);` ` ` ` ` `// Traversing over all bits of N` ` ` `for` `(` `int` `i = 0; i < MSB; i++) {` ` ` ` ` `// If ith bit of N is 0` ` ` `if` `((N & (1 << i))==0) {` ` ` ` ` `// Set ith bit of X to 1` ` ` `X += 1 << i;` ` ` ` ` `// Set ith bit of Y to 1` ` ` `Y += 1 << i;` ` ` `}` ` ` `}` ` ` ` ` `// Print Answer` ` ` `Console.Write(X+` `" "` `+Y);` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `N = 45;` ` ` `maximizeProduct(N);` ` ` `}` `}` `// This code is contributed by code_hunt.` |

## Javascript

`<script>` ` ` `// JavaScript Program to implement` ` ` `// the above approach` ` ` `// Function to find the pair (X, Y) such` ` ` `// that X xor Y = N and the count of set` ` ` `// bits in X and Y is less than count of` ` ` `// set bit in N` ` ` `function` `maximizeProduct(N)` ` ` `{` ` ` ` ` `// Stores MSB (Most Significant Bit)` ` ` `let MSB = Math.log2(N);` ` ` `// Stores the value of X` ` ` `let X = 1 << MSB;` ` ` `// Stores the value of Y` ` ` `let Y = N - (1 << MSB);` ` ` `// Traversing over all bits of N` ` ` `for` `(let i = 0; i < MSB; i++) {` ` ` `// If ith bit of N is 0` ` ` `if` `(!(N & (1 << i))) {` ` ` `// Set ith bit of X to 1` ` ` `X += 1 << i;` ` ` `// Set ith bit of Y to 1` ` ` `Y += 1 << i;` ` ` `}` ` ` `}` ` ` `// Print Answer` ` ` `document.write(X + ` `" "` `+ Y);` ` ` `}` ` ` `// Driver Code` ` ` `let N = 45;` ` ` `maximizeProduct(N);` `// This code is contributed by Potta Lokesh` ` ` `</script>` |

**Output:**

50 31

**Time Complexity:** O(log N)**Auxiliary Space:** O(N)