Skip to content
Related Articles

Related Articles

Maximize the Expression | Bit Manipulation
  • Difficulty Level : Medium
  • Last Updated : 26 Apr, 2021

Given two positive integers A and B. Let’s define D such that B AND D = D. The task is to maximize the expression A XOR D.
Examples: 
 

Input: A = 11 B = 4
Output: 15
Take D = 4 as (B AND D) = (4 AND 4) = 4.
Also, (A XOR D) = (11 XOR 4) = 15 which is the 
maximum according to the given condition.

Input: A = 9 and B = 13
Output: 13

 

Naive approach: Since B AND D = D, D will always be smaller than or equal to B. Hence, one can run a loop from 1 to B and check whether the given conditions are satisfied or not.
Efficient approach: Instead of running a loop and checking for each D, the maximum value of the expression (A XOR D) can be easily calculated using Bit Manipulation techniques. 
Let’s take an example to understand the way to approach the problem: 
 

A = 11 = 1011, B = 14 = 1110
Let's assume D = abcd in base 2 notation

B AND D:     1110           A XOR D:     1011
             abcd                        abcd   
            ------                      ------
             abcd                        ????

At 0th place: (0 AND d) = d implies d = 0 
At 1st place: (1 AND c) = c implies c = 0, 1 but to maximize (A XOR D), take c = 0
At 2nd place: (1 AND b) = b implies b = 0, 1 but to maximize (A XOR D), take b = 1
At 3rd place: (1 AND a) = a implies a = 0, 1 but to maximize (A XOR D), take a = 0

Hence, D = 0100 = 4 and maximum value of (A XOR D) = (11 XOR 4) = 15.

Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <iostream>
using namespace std;
 
#define MAX 32
 
// Function to return the value of
// the maximized expression
int maximizeExpression(int a, int b)
{
    int result = a;
 
    // int can have 32 bits
    for (int bit = MAX - 1; bit >= 0; bit--) {
 
        // Consider the ith bit of D to be 1
        int bitOfD = 1 << bit;
 
        // Calculate the value of (B AND bitOfD)
        int x = b & bitOfD;
 
        // Check if bitOfD satisfies (B AND D = D)
        if (x == bitOfD) {
 
            // Check if bitOfD can maximize (A XOR D)
            int y = result & bitOfD;
            if (y == 0) {
                result = result ^ bitOfD;
            }
        }
 
        // Note that we do not need to consider ith bit of D
        // to be 0 because if above condition are not satisfied
        // then value of result will not change
        // which is similar to considering bitOfD = 0
        // as result XOR 0 = result
    }
 
    return result;
}
 
// Driver code
int main()
{
    int a = 11, b = 14;
 
    cout << maximizeExpression(a, b);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
    final static int MAX = 32;
     
    // Function to return the value of
    // the maximized expression
    static int maximizeExpression(int a, int b)
    {
        int result = a;
     
        // int can have 32 bits
        for (int bit = MAX - 1; bit >= 0; bit--)
        {
     
            // Consider the ith bit of D to be 1
            int bitOfD = 1 << bit;
     
            // Calculate the value of (B AND bitOfD)
            int x = b & bitOfD;
     
            // Check if bitOfD satisfies (B AND D = D)
            if (x == bitOfD) {
     
                // Check if bitOfD can maximize (A XOR D)
                int y = result & bitOfD;
                if (y == 0)
                {
                    result = result ^ bitOfD;
                }
            }
     
            // Note that we do not need to consider ith bit of D
            // to be 0 because if above condition are not satisfied
            // then value of result will not change
            // which is similar to considering bitOfD = 0
            // as result XOR 0 = result
        }
        return result;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int a = 11, b = 14;
     
        System.out.println(maximizeExpression(a, b));
    }
}
 
// This code is contributed by AnkitRai01

Python3




# Python3 implementation of the approach
MAX = 32
 
# Function to return the value of
# the maximized expression
def maximizeExpression(a, b) :
 
    result = a
 
    # int can have 32 bits
    for bit in range(MAX - 1, -1, -1) :
 
        # Consider the ith bit of D to be 1
        bitOfD = 1 << bit
 
        # Calculate the value of (B AND bitOfD)
        x = b & bitOfD
 
        # Check if bitOfD satisfies (B AND D = D)
        if (x == bitOfD) :
 
            # Check if bitOfD can maximize (A XOR D)
            y = result & bitOfD
            if (y == 0) :
                result = result ^ bitOfD
 
        # Note that we do not need to consider ith bit of D
        # to be 0 because if above condition are not satisfied
        # then value of result will not change
        # which is similar to considering bitOfD = 0
        # as result XOR 0 = result
 
    return result
 
# Driver code
a = 11
b = 14
print(maximizeExpression(a, b))
 
# This code is contributed by divyamohan123

C#




// C# implementation of the above approach
using System;
class GFG
{
    static int MAX = 32;
     
    // Function to return the value of
    // the maximized expression
    static int maximizeExpression(int a, int b)
    {
        int result = a;
     
        // int can have 32 bits
        for (int bit = MAX - 1; bit >= 0; bit--)
        {
     
            // Consider the ith bit of D to be 1
            int bitOfD = 1 << bit;
     
            // Calculate the value of (B AND bitOfD)
            int x = b & bitOfD;
     
            // Check if bitOfD satisfies (B AND D = D)
            if (x == bitOfD)
            {
     
                // Check if bitOfD can maximize (A XOR D)
                int y = result & bitOfD;
                if (y == 0)
                {
                    result = result ^ bitOfD;
                }
            }
     
            // Note that we do not need to consider
            // ith bit of D to be 0 because if
            // above condition are not satisfied then
            // value of result will not change which is
            // similar to considering bitOfD = 0 as
            // result XOR 0 = result
        }
        return result;
    }
     
    // Driver code
    public static void Main (String []args)
    {
        int a = 11, b = 14;
     
        Console.WriteLine(maximizeExpression(a, b));
    }
}
 
// This code is contributed by Arnab Kundu

Javascript




<script>
 
// Javascript implementation of the approach
 
let MAX = 32;
 
// Function to return the value of
// the maximized expression
function maximizeExpression(a, b)
{
    let result = a;
 
    // int can have 32 bits
    for (let bit = MAX - 1; bit >= 0; bit--)
    {
 
        // Consider the ith bit of D to be 1
        let bitOfD = 1 << bit;
 
        // Calculate the value of (B AND bitOfD)
        let x = b & bitOfD;
 
        // Check if bitOfD satisfies (B AND D = D)
        if (x == bitOfD) {
 
            // Check if bitOfD can maximize (A XOR D)
            let y = result & bitOfD;
            if (y == 0) {
                result = result ^ bitOfD;
            }
        }
 
        // Note that we do not need
        // to consider ith bit of D
        // to be 0 because if above
        // condition are not satisfied
        // then value of result will not change
        // which is similar to considering bitOfD = 0
        // as result XOR 0 = result
    }
 
    return result;
}
 
// Driver code
    let a = 11, b = 14;
 
    document.write(maximizeExpression(a, b));
 
</script>
Output: 
15

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live and Geeks Classes Live USA




My Personal Notes arrow_drop_up
Recommended Articles
Page :