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

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++ 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;
    }

    // Simply return the required maximum value.
    return (1 << msbPos) -1;   // 2 ^ msbPos - 1
}

//  Driver code to test above methods
int main()
{
    int L = 8;
    int R = 20;
    cout << maxXORInRange(L, R) << endl;
    return 0;
}

// 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 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# 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 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
?>

<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

Time Complexity: O(log(R))

Auxiliary Space: O(1)