Value in a given range with maximum XOR
Given positive integers N, L, and R, we have to find the maximum value of N ? X, where X ? [L, R].
Examples:
Input : N = 7
L = 2
R = 23
Output : 23
Explanation : When X = 16, we get 7 ? 16 = 23 which is the maximum value for all X ? [2, 23].
Input : N = 10
L = 5
R = 12
Output : 15
Explanation : When X = 5, we get 10 ? 5 = 15 which is the maximum value for all X ? [5, 12].
Brute force approach: We can solve this problem using brute force approach by looping over all integers over the range [L, R] and taking their XOR with N while keeping a record of the maximum result encountered so far. The complexity of this algorithm will be O(R – L), and it is not feasible when the input variables approach high values such as 109.
Efficient approach: Since the XOR of two bits is 1 if and only if they are complementary to each other, we need X to have complementary bits to that of N to have the maximum value. We will iterate from the largest bit (log2(R)th Bit) to the lowest (0th Bit). The following two cases can arise for each bit:
- If the bit is not set, i.e. 0, we will try to set it in X. If setting this bit to 1 results in X exceeding R, then we will not set it.
- If the bit is set, i.e. 1, then we will try to unset it in X. If the current value of X is already greater than or equal to L, then we can safely unset the bit. In the other case, we will check if setting all of the next bits is enough to keep X >= L. If not, then we are required to set the current bit. Observe that setting all the next bits is equivalent to adding (1 << b) – 1, where b is the current bit.
The time complexity of this approach is O(log2(R)).
C++
#include <cmath>
#include <iostream>
using namespace std;
int maximumXOR( int n, int l, int r)
{
int x = 0;
for ( int i = log2(r); i >= 0; --i)
{
if (n & (1 << i))
{
if (x + (1 << i) - 1 < l)
x ^= (1 << i);
}
else
{
if ((x ^ (1 << i)) <= r)
x ^= (1 << i);
}
}
return n ^ x;
}
int main()
{
int n = 7, l = 2, r = 23;
cout << "The output is " << maximumXOR(n, l, r);
return 0;
}
|
Java
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
static int maximumXOR( int n, int l, int r)
{
int x = 0 ;
for ( int i = ( int )(Math.log(r)/Math.log( 2 )); i >= 0 ; --i)
{
if ((n & ( 1 << i))> 0 )
{
if (x + ( 1 << i) - 1 < l)
x ^= ( 1 << i);
}
else
{
if ((x ^ ( 1 << i)) <= r)
x ^= ( 1 << i);
}
}
return n ^ x;
}
public static void main(String args[])
{
int n = 7 , l = 2 , r = 23 ;
System.out.println( "The output is " + maximumXOR(n, l, r));
}
}
|
Python3
import math
def maximumXOR(n, l, r):
x = 0
for i in range ( int (math.log2(r)), - 1 , - 1 ):
if (n & ( 1 << i)):
if (x + ( 1 << i) - 1 < l):
x ^ = ( 1 << i)
else :
if (x ^ ( 1 << i)) < = r:
x ^ = ( 1 << i)
return n ^ x
n = 7
l = 2
r = 23
print ( "The output is" ,
maximumXOR(n, l, r))
|
C#
using System;
class GFG
{
public static int maximumXOR( int n,
int l, int r)
{
int x = 0;
for ( int i = ( int )(Math.Log(r) /
Math.Log(2)); i >= 0; --i)
{
if ((n & (1 << i)) > 0)
{
if (x + (1 << i) - 1 < l)
{
x ^= (1 << i);
}
}
else
{
if ((x ^ (1 << i)) <= r)
{
x ^= (1 << i);
}
}
}
return n ^ x;
}
public static void Main( string [] args)
{
int n = 7, l = 2, r = 23;
Console.WriteLine( "The output is " +
maximumXOR(n, l, r));
}
}
|
Javascript
<script>
function maximumXOR(n, l, r)
{
let x = 0;
for (let i =
parseInt(Math.log(r) / Math.log(2)); i >= 0; --i)
{
if (n & (1 << i))
{
if (x + (1 << i) - 1 < l)
x ^= (1 << i);
}
else
{
if ((x ^ (1 << i)) <= r)
x ^= (1 << i);
}
}
return n ^ x;
}
let n = 7, l = 2, r = 23;
document.write( "The output is " + maximumXOR(n, l, r));
</script>
|
PHP
<?php
function maximumXOR( $n , $l , $r )
{
$x = 0;
for ( $i = log( $r , 2); $i >= 0; -- $i )
{
if ( $n & (1 << $i ))
{
if ( $x + (1 << $i ) - 1 < $l )
$x ^= (1 << $i );
}
else
{
if (( $x ^ (1 << $i )) <= $r )
$x ^= (1 << $i );
}
}
return $n ^ $x ;
}
$n = 7;
$l = 2;
$r = 23;
echo "The output is " ,
maximumXOR( $n , $l , $r );
?>
|
Time complexity: O(log2r)
Auxiliary Space: O(1)
Approach#2: Using Brute Force
One way to solve this problem is to try all possible values of X in the given range and find the one that gives the maximum XOR value with N.
Algorithm
1. Define a function max_XOR(N, L, R) that takes N, L and R as input.
2. Initialize a variable max_XOR_val to 0.
3. For each value of X in the range [L, R], calculate the XOR value of N and X.
4. If the XOR value is greater than max_XOR_val, update max_XOR_val with this value.
5. Return max_XOR_val as the output.
C++
#include <iostream>
using namespace std;
int max_XOR( int N, int L, int R) {
int max_XOR_val = 0;
for ( int X = L; X <= R; X++) {
int XOR_val = N ^ X;
if (XOR_val > max_XOR_val) {
max_XOR_val = XOR_val;
}
}
return max_XOR_val;
}
int main() {
int N = 7;
int L = 2;
int R = 23;
cout << max_XOR(N, L, R) << endl;
N = 10;
L = 5;
R = 12;
cout << max_XOR(N, L, R) << endl;
return 0;
}
|
Java
public class Main {
public static int maxXOR( int N, int L, int R) {
int maxXORValue = 0 ;
for ( int X = L; X <= R; X++) {
int XORValue = N ^ X;
if (XORValue > maxXORValue) {
maxXORValue = XORValue;
}
}
return maxXORValue;
}
public static void main(String[] args) {
int N = 7 ;
int L = 2 ;
int R = 23 ;
System.out.println( maxXOR(N, L, R));
N = 10 ;
L = 5 ;
R = 12 ;
System.out.println( maxXOR(N, L, R));
}
}
|
Python3
def max_XOR(N, L, R):
max_XOR_val = 0
for X in range (L, R + 1 ):
XOR_val = N ^ X
if XOR_val > max_XOR_val:
max_XOR_val = XOR_val
return max_XOR_val
N = 7
L = 2
R = 23
print (max_XOR(N, L, R))
N = 10
L = 5
R = 12
print (max_XOR(N, L, R))
|
C#
using System;
public class MainClass
{
public static int MaxXOR( int N, int L, int R)
{
int maxXORValue = 0;
for ( int X = L; X <= R; X++)
{
int XORValue = N ^ X;
if (XORValue > maxXORValue)
{
maxXORValue = XORValue;
}
}
return maxXORValue;
}
public static void Main( string [] args)
{
int N = 7;
int L = 2;
int R = 23;
Console.WriteLine(MaxXOR(N, L, R));
N = 10;
L = 5;
R = 12;
Console.WriteLine(MaxXOR(N, L, R));
}
}
|
Javascript
function max_XOR(N, L, R) {
let max_XOR_val = 0;
for (let X = L; X <= R; X++) {
let XOR_val = N ^ X;
if (XOR_val > max_XOR_val) {
max_XOR_val = XOR_val;
}
}
return max_XOR_val;
}
let N = 7;
let L = 2;
let R = 23;
console.log(max_XOR(N, L, R));
N = 10;
L = 5;
R = 12;
console.log(max_XOR(N, L, R));
|
Time Complexity: O(R-L+1)
Space Complexity: O(1)
Last Updated :
16 Oct, 2023
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...