# Maximum XOR value of a pair from a range

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

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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;` `}` |

## 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));` ` ` `}` `}` |

## 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.` |

## 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.` |

## 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` `?>` |

## Javascript

`<script>` ` ` `// Javascript 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` ` ` `let LXR = L ^ R;` ` ` ` ` `// loop to get msb position of L^R` ` ` `let msbPos = 0;` ` ` `while` `(LXR > 0)` ` ` `{` ` ` `msbPos++;` ` ` `LXR >>= 1;` ` ` `}` ` ` ` ` `// construct result by` ` ` `// adding 1, msbPos times` ` ` `let maxXOR = 0;` ` ` `let two = 1;` ` ` `while` `(msbPos-- > 0)` ` ` `{` ` ` `maxXOR += two;` ` ` `two <<= 1;` ` ` `}` ` ` ` ` `return` `maxXOR;` ` ` `}` ` ` ` ` `let L = 8;` ` ` `let R = 20;` ` ` `document.write(maxXORInRange(L, R));` ` ` `</script>` |

**Output :**

31

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