Given a natural number** N**, the task is to find the largest number **M** having the same length in binary representation as **N** such that the difference between **N |** **M **and **N ^ M** is maximum.

**Examples:**

Input:N = 6Output:7Explanation:

All number numbers having same length in binary representation as N are 4, 5, 6, 7.

(6 | 4) – (6 ^ 4) = 4

(6 | 5) – (6 ^ 5) = 4

(6 | 6) – (6 ^ 6) = 6

(6 | 7) – (6 ^ 7) = 6

Hence, largest M for which (N | M) – (N ^ M) is maximum is 7

Input:N = 10Output:15Explanation:

The largest number M = 15 which has the same length in binary representation as 10 and the difference betweenN |MandN ^ Mis maximum.

**Naive Approach:** The idea is to simply find all the numbers having the same length in binary representation as** N** and then for every number iterate and find the largest** **integer** **having **(N | i) – (N ^ i) **maximum.

**Time Complexity:** O(N)**Auxiliary Space:** O(1)

**Efficient Approach:** The idea is to initialize **M = 0** and iterate bit by bit in N (say **i**) and set or unset the** i ^{th}** bit of

**M**according to the following 2 observations :

- When an i
^{th}bit of N is**set**: In this case, if we**unset**the i^{th}bit of M, i^{th}bit of both**N | M**and**N^M**will be set whereas on setting this bit of M, an i^{th}bit of**N|M**will be set and**N^M**will be unset which will increase**(N | M) – (N ^ M)**. Hence, it is optimal to set this bit of**M**. - When an i
^{th}bit of N is**unset:**In this case, if we**set**this bit of**N|M**and N^M will have this bit set or on keeping this bit of M unset both**N|M**and**N^M**will have this bit unset. So, in this case, we cannot increase the difference between them but as the requirement is to output the maximum**M**possible, so set this bit of**M**. - From the above observations, it is clear that
**M**will have all the bits set.

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 largest number` `// M having the same length in binary` `// form as N such that the difference` `// between N | M and N ^ M is maximum` `int` `maxORminusXOR(` `int` `N)` `{` ` ` `// Find the most significant` ` ` `// bit of N` ` ` `int` `MSB = log2(N);` ` ` `// Initialize M` ` ` `int` `M = 0;` ` ` `// Set all the bits of M` ` ` `for` `(` `int` `i = 0; i <= MSB; i++)` ` ` `M += (1 << i);` ` ` `// Return the answer` ` ` `return` `M;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given Number N` ` ` `int` `N = 10;` ` ` `// Function Call` ` ` `cout << maxORminusXOR(N);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to find the largest number` `// M having the same length in binary` `// form as N such that the difference` `// between N | M and N ^ M is maximum` `static` `int` `maxORminusXOR(` `int` `N)` `{` ` ` ` ` `// Find the most significant` ` ` `// bit of N` ` ` `int` `MSB = (` `int` `)Math.ceil(Math.log(N));` ` ` `// Initialize M` ` ` `int` `M = ` `0` `;` ` ` `// Set all the bits of M` ` ` `for` `(` `int` `i = ` `0` `; i <= MSB; i++)` ` ` `M += (` `1` `<< i);` ` ` `// Return the answer` ` ` `return` `M;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` ` ` `// Given number N` ` ` `int` `N = ` `10` `;` ` ` `// Function call` ` ` `System.out.print(maxORminusXOR(N));` `}` `}` `// This code is contributed by Rajput-Ji` |

## Python3

`# Python3 program for the above approach` `import` `math` `# Function to find the largest number` `# M having the same length in binary` `# form as N such that the difference` `# between N | M and N ^ M is maximum` `def` `maxORminusXOR(N):` ` ` `# Find the most significant` ` ` `# bit of N` ` ` `MSB ` `=` `int` `(math.log2(N));` ` ` `# Initialize M` ` ` `M ` `=` `0` ` ` `# Set all the bits of M` ` ` `for` `i ` `in` `range` `(MSB ` `+` `1` `):` ` ` `M ` `+` `=` `(` `1` `<< i)` ` ` `# Return the answer` ` ` `return` `M` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `# Given Number N` ` ` `N ` `=` `10` ` ` `# Function call` ` ` `print` `(maxORminusXOR(N))` `# This code is contributed by jana_sayantan` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG{` `// Function to find the largest number` `// M having the same length in binary` `// form as N such that the difference` `// between N | M and N ^ M is maximum` `static` `int` `maxORminusXOR(` `int` `N)` `{` ` ` ` ` `// Find the most significant` ` ` `// bit of N` ` ` `int` `MSB = (` `int` `)Math.Ceiling(Math.Log(N));` ` ` `// Initialize M` ` ` `int` `M = 0;` ` ` `// Set all the bits of M` ` ` `for` `(` `int` `i = 0; i <= MSB; i++)` ` ` `M += (1 << i);` ` ` `// Return the answer` ` ` `return` `M;` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` ` ` `// Given number N` ` ` `int` `N = 10;` ` ` `// Function call` ` ` `Console.Write(maxORminusXOR(N));` `}` `}` `// This code is contributed by 29AjayKumar` |

## Javascript

`<script>` `// JavaScript implementation of the above approach` ` ` `// Function to find the largest number` `// M having the same length in binary` `// form as N such that the difference` `// between N | M and N ^ M is maximum` `function` `maxORminusXOR(N)` `{` ` ` ` ` `// Find the most significant` ` ` `// bit of N` ` ` `let MSB = Math.ceil(Math.log(N));` ` ` ` ` `// Initialize M` ` ` `let M = 0;` ` ` ` ` `// Set all the bits of M` ` ` `for` `(let i = 0; i <= MSB; i++)` ` ` `M += (1 << i);` ` ` ` ` `// Return the answer` ` ` `return` `M;` `}` `// Driver code` ` ` ` ` `// Given number N` ` ` `let N = 10;` ` ` ` ` `// Function call` ` ` `document.write(maxORminusXOR(N));` ` ` ` ` `// This code is contributed by code_hunt.` `</script>` |

**Output:**

15

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

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