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)