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 (0^th Bit). The following two cases can arise for each bit: 

1. 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. 
    
2. 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)).

// CPP program to find the x in range [l, r]
// such that x ^ n is maximum.
#include <cmath>
#include <iostream>
using namespace std;

// Function to calculate the maximum value of
// N ^ X, where X is in the range [L, R]
int maximumXOR(int n, int l, int r)
{
    int x = 0;
    for (int i = log2(r); i >= 0; --i) 
    {
        if (n & (1 << i))  // Set bit
        { 
            if (x + (1 << i) - 1 < l)
                x ^= (1 << i);
        }
        else // Unset bit
        { 
            if ((x ^ (1 << i)) <= r)
                x ^= (1 << i);
        }
    }
    return n ^ x;
}

// Driver Code
int main()
{
    int n = 7, l = 2, r = 23;
    cout << "The output is " << maximumXOR(n, l, r);
    return 0;
}

// Java program to find the x in range [l, r] 
// such that x ^ n is maximum.

import java.util.*;
import java.lang.*;
import java.io.*;

class GFG
{
// Function to calculate the maximum value of 
// N ^ X, where X is in the range [L, R]
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) // Set bit
        {
            if  (x + (1 << i) - 1 < l)
                x ^= (1 << i);
        }
        else // Unset bit
        {
            if ((x ^ (1 << i)) <= r)
                x ^= (1 << i);
        }
    }
    return n ^ x;
}

// Driver function
public static void main(String args[])
{
    int n = 7, l = 2, r = 23;
    System.out.println( "The output is " + maximumXOR(n, l, r));

}
}

// This code is Contributed by tufan_gupta2000

# Python program to find the 
# x in range [l, r] such that
# x ^ n is maximum. 
import math

# Function to calculate the 
# maximum value of N ^ X, 
# where X is in the range [L, R] 
def maximumXOR(n, l, r):
    x = 0
    for i in range(int(math.log2(r)), -1, -1): 
        if (n & (1 << i)): # Set bit 
            if  (x + (1 << i) - 1 < l):
                x ^= (1 << i)
        else: # Unset bit
            if (x ^ (1 << i)) <= r:
                x ^= (1 << i)
    return n ^ x

# Driver code
n = 7
l = 2
r = 23
print("The output is", 
       maximumXOR(n, l, r))

# This code was contributed
# by VishalBachchas

// C# program to find the x in range 
// [l, r] such that x ^ n is maximum. 
using System;

class GFG
{
    
// Function to calculate the 
// maximum value of N ^ X,
// where X is in the range [L, R] 
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) // Set bit
        {
            if (x + (1 << i) - 1 < l)
            {
                x ^= (1 << i);
            }
        }
        else // Unset bit
        {
            if ((x ^ (1 << i)) <= r)
            {
                x ^= (1 << i);
            }
        }
    }
    return n ^ x;
}

// Driver Code 
public static void Main(string[] args)
{
    int n = 7, l = 2, r = 23;
    Console.WriteLine("The output is " + 
                   maximumXOR(n, l, r));
}
}

// This code is contributed
// by Shrikant13

<script>

// Javascript program to find 
// the x in range [l, r]
// such that x ^ n is maximum.

// Function to calculate the maximum value of
// N ^ X, where X is in the range [L, R]
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))  // Set bit
        { 
            if (x + (1 << i) - 1 < l)
                x ^= (1 << i);
        }
        else // Unset bit
        { 
            if ((x ^ (1 << i)) <= r)
                x ^= (1 << i);
        }
    }
    return n ^ x;
}

// Driver Code
    let n = 7, l = 2, r = 23;
    document.write("The output is " + maximumXOR(n, l, r));
    
</script>

<?php
// PHP program to find the x in range 
// [l, r] such that x ^ n is maximum. 

// Function to calculate the maximum 
// value of N ^ X, where X is in the
// range [L, R] 
function maximumXOR($n, $l, $r) 
{ 
    $x = 0; 
    for ($i = log($r, 2); $i >= 0; --$i)
    { 
        if ($n & (1 << $i)) 
        {   
            // Set bit 
            if ($x + (1 << $i) - 1 < $l)
                $x ^= (1 << $i); 
        } 
        else
        {
            // Unset bit 
            if (($x ^ (1 << $i)) <= $r) 
                $x ^= (1 << $i); 
        } 
    } 
    return $n ^ $x; 
} 

// Driver Code 
$n = 7;
$l = 2;
$r = 23; 
echo "The output is " , 
      maximumXOR($n, $l, $r); 

// This code is contributed by ajit
?>

The output is 23

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.

#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;
}

public class Main {

    // Function to find the maximum XOR value between N and integers in the range [L, R]
    public static int maxXOR(int N, int L, int R) {
        int maxXORValue = 0;
        for (int X = L; X <= R; X++) {
            int XORValue = N ^ X; // Calculate the XOR value between N and X
            if (XORValue > maxXORValue) {
                maxXORValue = XORValue; // Update the maximum XOR value if a larger one is found
            }
        }
        return maxXORValue;
    }

    public static void main(String[] args) {
        int N = 7; 
        int L = 2; 
        int R = 23; 
        
        // Find and print the maximum XOR value for the given parameters
        System.out.println( maxXOR(N, L, R));

        N = 10; 
        L = 5; 
        R = 12;
        
        // Find and print the maximum XOR value for the updated parameters
        System.out.println( maxXOR(N, L, R));
    }
}

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

# Example usage
N = 7
L = 2
R = 23
print(max_XOR(N, L, R)) 

N = 10
L = 5
R = 12
print(max_XOR(N, L, R)) 

using System;

public class MainClass
{
    // Function to find the maximum XOR value between N and integers in the range [L, R]
    public static int MaxXOR(int N, int L, int R)
    {
        int maxXORValue = 0;
        for (int X = L; X <= R; X++)
        {
            int XORValue = N ^ X; // Calculate the XOR value between N and X
            if (XORValue > maxXORValue)
            {
                maxXORValue = XORValue; // Update the maximum XOR value if a larger one is found
            }
        }
        return maxXORValue;
    }

    public static void Main(string[] args)
    {
        int N = 7;
        int L = 2;
        int R = 23;

        // Find and print the maximum XOR value for the given parameters
        Console.WriteLine(MaxXOR(N, L, R));

        N = 10;
        L = 5;
        R = 12;

        // Find and print the maximum XOR value for the updated parameters
        Console.WriteLine(MaxXOR(N, L, R));
    }
}

// Function to find the maximum XOR value between N and numbers in the range [L, R]
function max_XOR(N, L, R) {
    // Variable to store the maximum XOR value
    let max_XOR_val = 0;

    // Iterate over the range [L, R]
    for (let X = L; X <= R; X++) {
        // Calculate the XOR value between N and X
        let XOR_val = N ^ X;

        // Update the maximum XOR value if necessary
        if (XOR_val > max_XOR_val) {
            max_XOR_val = XOR_val;
        }
    }

    // Return the maximum XOR value
    return max_XOR_val;
}

// Example usage
let N = 7;
let L = 2;
let R = 23;
console.log(max_XOR(N, L, R)); // Output: 31

N = 10;
L = 5;
R = 12;
console.log(max_XOR(N, L, R)); // Output: 15

23
15

Time Complexity: O(R-L+1)
Space Complexity: O(1)

vaibhav29498

Follow

Improve

Article Tags :

• Bit Magic
• Greedy
• DSA
• Bitwise-XOR

Practice Tags :

• Bit Magic
• Greedy