Maximum XOR value of a pair from a range - GeeksforGeeks

05 Jan, 2022
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++
Java
Python3
C#
PHP
Javascript
// 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 (pow(2, msbPos)-1);} //  Driver code to test above methodsint 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 codeL, R = 8, 20print(maxXORInRange(L, R)) # This code is contributed by Anant Agarwal.
// C# program to get maximum xor// value of two numbers in a rangeusing 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
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