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Maximum XOR value of a pair from a range
• Difficulty Level : Medium
• Last Updated : 12 Sep, 2018

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ``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

 `>= 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``?>`

Output :
```31
```

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