Given a range [L, R], we need to find two integers in this range such that their XOR is maximum among all possible choices of two integers. More Formally,

given [L, R], find max (A ^ B) where L <= A, B

**Examples :**

Input : L = 8 R = 20 Output : 31 31 is XOR of 15 and 16. Input : L = 1 R = 3 Output : 3

A **simple solution** is to generate all pairs, find their XOR values and finally return the maximum XOR value.

An **efficient solution **is to consider pattern of binary values from L to R. We can see that first bit from L to R either changes from 0 to 1 or it stays 1 i.e. if we take the XOR of any two numbers for maximum value their first bit will be fixed which will be same as first bit of XOR of L and R itself.

After observing the technique to get first bit, we can see that if we XOR L and R, the most significant bit of this XOR will tell us the maximum value we can achieve i.e. let XOR of L and R is 1xxx where x can be 0 or 1 then maximum XOR value we can get is 1111 because from L to R we have all possible combination of xxx and it is always possible to choose these bits in such a way from two numbers such that their XOR becomes all 1. It is explained below with some examples,

Examples 1: L = 8 R = 20 L ^ R = (01000) ^ (10100) = (11100) Now as L ^ R is of form (1xxxx) we can get maximum XOR as (11111) by choosing A and B as 15 and 16 (01111 and 10000) Examples 2: L = 16 R = 20 L ^ R = (10000) ^ (10100) = (00100) Now as L ^ R is of form (1xx) we can get maximum xor as (111) by choosing A and B as 19 and 20 (10011 and 10100)

So the solution of this problem depends on the value of (L ^ R) only. We will calculate the L^R value first and then from most significant bit of this value, we will add all 1s to get the final result.

## C++

`// C/C++ program to get maximum xor value ` `// of two numbers in a range ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// method to get maximum xor value in range [L, R] ` `int` `maxXORInRange(` `int` `L, ` `int` `R) ` `{ ` ` ` `// get xor of limits ` ` ` `int` `LXR = L ^ R; ` ` ` ` ` `// loop to get msb position of L^R ` ` ` `int` `msbPos = 0; ` ` ` `while` `(LXR) ` ` ` `{ ` ` ` `msbPos++; ` ` ` `LXR >>= 1; ` ` ` `} ` ` ` ` ` `// construct result by adding 1, ` ` ` `// msbPos times ` ` ` `int` `maxXOR = 0; ` ` ` `int` `two = 1; ` ` ` `while` `(msbPos--) ` ` ` `{ ` ` ` `maxXOR += two; ` ` ` `two <<= 1; ` ` ` `} ` ` ` ` ` `return` `maxXOR; ` `} ` ` ` `// Driver code to test above methods ` `int` `main() ` `{ ` ` ` `int` `L = 8; ` ` ` `int` `R = 20; ` ` ` `cout << maxXORInRange(L, R) << endl; ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to get maximum xor value ` `// of two numbers in a range ` ` ` `class` `Xor ` `{ ` ` ` `// method to get maximum xor value in range [L, R] ` ` ` `static` `int` `maxXORInRange(` `int` `L, ` `int` `R) ` ` ` `{ ` ` ` `// get xor of limits ` ` ` `int` `LXR = L ^ R; ` ` ` ` ` `// loop to get msb position of L^R ` ` ` `int` `msbPos = ` `0` `; ` ` ` `while` `(LXR > ` `0` `) ` ` ` `{ ` ` ` `msbPos++; ` ` ` `LXR >>= ` `1` `; ` ` ` `} ` ` ` ` ` `// construct result by adding 1, ` ` ` `// msbPos times ` ` ` `int` `maxXOR = ` `0` `; ` ` ` `int` `two = ` `1` `; ` ` ` `while` `(msbPos-- >` `0` `) ` ` ` `{ ` ` ` `maxXOR += two; ` ` ` `two <<= ` `1` `; ` ` ` `} ` ` ` ` ` `return` `maxXOR; ` ` ` `} ` ` ` ` ` `// main function ` ` ` `public` `static` `void` `main (String[] args) ` ` ` `{ ` ` ` `int` `L = ` `8` `; ` ` ` `int` `R = ` `20` `; ` ` ` `System.out.println(maxXORInRange(L, R)); ` ` ` `} ` `} ` |

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## Python3

`# Python3 program to get maximum xor ` `# value of two numbers in a range ` ` ` `# Method to get maximum xor ` `# value in range [L, R] ` `def` `maxXORInRange(L, R): ` ` ` ` ` `# get xor of limits ` ` ` `LXR ` `=` `L ^ R ` ` ` ` ` `# loop to get msb position of L^R ` ` ` `msbPos ` `=` `0` ` ` `while` `(LXR): ` ` ` ` ` `msbPos ` `+` `=` `1` ` ` `LXR >>` `=` `1` ` ` ` ` ` ` `# construct result by adding 1, ` ` ` `# msbPos times ` ` ` `maxXOR, two ` `=` `0` `, ` `1` ` ` ` ` `while` `(msbPos): ` ` ` ` ` `maxXOR ` `+` `=` `two ` ` ` `two <<` `=` `1` ` ` `msbPos ` `-` `=` `1` ` ` ` ` `return` `maxXOR ` ` ` `# Driver code ` `L, R ` `=` `8` `, ` `20` `print` `(maxXORInRange(L, R)) ` ` ` `# This code is contributed by Anant Agarwal. ` |

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## C#

`// C# program to get maximum xor ` `// value of two numbers in a range ` `using` `System; ` ` ` `class` `Xor ` `{ ` ` ` ` ` `// method to get maximum xor ` ` ` `// value in range [L, R] ` ` ` `static` `int` `maxXORInRange(` `int` `L, ` `int` `R) ` ` ` `{ ` ` ` ` ` `// get xor of limits ` ` ` `int` `LXR = L ^ R; ` ` ` ` ` `// loop to get msb position of L^R ` ` ` `int` `msbPos = 0; ` ` ` `while` `(LXR > 0) ` ` ` `{ ` ` ` `msbPos++; ` ` ` `LXR >>= 1; ` ` ` `} ` ` ` ` ` `// construct result by ` ` ` `// adding 1, msbPos times ` ` ` `int` `maxXOR = 0; ` ` ` `int` `two = 1; ` ` ` `while` `(msbPos-- >0) ` ` ` `{ ` ` ` `maxXOR += two; ` ` ` `two <<= 1; ` ` ` `} ` ` ` ` ` `return` `maxXOR; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `L = 8; ` ` ` `int` `R = 20; ` ` ` `Console.WriteLine(maxXORInRange(L, R)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Anant Agarwal. ` |

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## PHP

`<?php ` `// PHP program to get maximum ` `// xor value of two numbers ` `// in a range ` ` ` `// method to get maximum xor ` `// value in range [L, R] ` `function` `maxXORInRange(` `$L` `, ` `$R` `) ` `{ ` ` ` `// get xor of limits ` ` ` `$LXR` `= ` `$L` `^ ` `$R` `; ` ` ` ` ` `// loop to get msb ` ` ` `// position of L^R ` ` ` `$msbPos` `= 0; ` ` ` `while` `(` `$LXR` `) ` ` ` `{ ` ` ` `$msbPos` `++; ` ` ` `$LXR` `>>= 1; ` ` ` `} ` ` ` ` ` `// construct result by ` ` ` `// adding 1, msbPos times ` ` ` `$maxXOR` `= 0; ` ` ` `$two` `= 1; ` ` ` `while` `(` `$msbPos` `--) ` ` ` `{ ` ` ` `$maxXOR` `+= ` `$two` `; ` ` ` `$two` `<<= 1; ` ` ` `} ` ` ` ` ` `return` `$maxXOR` `; ` `} ` ` ` `// Driver Code ` `$L` `= 8; ` `$R` `= 20; ` `echo` `maxXORInRange(` `$L` `, ` `$R` `), ` `"\n"` `; ` ` ` `// This code is contributed by aj_36 ` `?> ` |

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**Output :**

31

This article is contributed by **Utkarsh Trivedi**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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