Maximum XOR value of a pair from a range

2.6

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

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 (0xx) 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));
    }
}


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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.

Recommended Posts:



2.6 Average Difficulty : 2.6/5.0
Based on 5 vote(s)










Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.