Maximum value of B less than A such that A ^ B = A + B

Given an integer A, the task is to find the maximum value possible(B) which is less than A, such that xor of these two numbers A and B is equal their sum, that is A ^ B = A + B.

Examples:  

Input: A = 4 
Output:
Explanation: 
There are many such integers such that A ^ B = A + B 
Some of this integers are – 
4 ^ 3 = 4 + 3 = 7 
4 ^ 2 = 4 + 2 = 6 
4 ^ 1 = 4 + 1 = 5 
4 ^ 0 = 4 + 0 = 4 
Maximum of these values is 3

Input:
Output:
There is no integer except 0 such that A + B = A ^ B 

Approach: The idea is to use the fact that 



A + B &= (A \wedge B) + 2*(A \& B)

and to get the value of 

A + B = A \wedge B

, the value of (A & B) must be equal to 0.  

=> A & B = 0
=> B = ~A

For Example:  

A = 4 (1 0 0)  
B = ~ A = (0 1 1) = 3 

Below is the implementation of the above approach: 

C++

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// C++ implementation to find
// maximum value of B such that
// A ^ B = A + B
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum
// value of B such that A^B = A+B
void maxValue(int a)
{
     
    // Binary Representation of A
    string c = bitset<3>(a).to_string();
    string b = "";
     
    // Loop to find the negation
    // of the integer A
    for(int i = 0; i < c.length(); i++)
    {
        if ((c[i] - '0') == 1)
            b += '0';
        else
            b += '1';
    }
        
    // Output
    cout << bitset<3>(b).to_ulong();
}
 
// Driver code
int main()
{
    int a = 4;
     
    // Function Call
    maxValue(a);
     
    return 0;
}
 
// This code is contributed by divyeshrabadiya07

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Java

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// Java implementation to find
// maximum value of B such that
// A ^ B = A + B
  
// Function to find the maximum
// value of B such that A^B = A+B
class GFG
{
 
static void maxValue(int a)
{
      
    // Binary Representation of A
    String c = Integer.toBinaryString(a);
     
    String b = "";
      
    // Loop to find the negation
    // of the integer A
    for (int i = 0; i < c.length(); i++)
    {
        if((c.charAt(i)-'0')==1)
            b +='0';
        else
            b+='1';
    }
      
    // output
    System.out.print(Integer.parseInt(b, 2));
    
}
  
// Driver Code
public static void main(String []args)
{
    int a = 4;
      
    // Function Call
    maxValue(a);
}
}
 
// This code is contributed by chitranayal

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Python3

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# Python3 implementation to find
# maximum value of B such that
# A ^ B = A + B
 
# Function to find the maximum
# value of B such that A^B = A+B
def maxValue(a):
     
    # Binary Representation of A
    a = bin(a)[2:]
     
    b = ''
     
    # Loop to find the negation
    # of the integer A
    for i in list(a):
        b += str(int(not int(i)))
         
    # output
    print(int(b, 2))
    return int(b, 2)
 
# Driver Code
if __name__ == '__main__':
    a = 4
     
    # Function Call
    maxValue(a)

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C#

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// C# implementation to find
// maximum value of B such that
// A ^ B = A + B
   
// Function to find the maximum
// value of B such that A^B = A+B
using System;
using System.Collections.Generic;
 
class GFG
{
  
static void maxValue(int a)
{
       
    // Binary Representation of A
    String c = Convert.ToString(a, 2);
      
    String b = "";
       
    // Loop to find the negation
    // of the integer A
    for (int i = 0; i < c.Length; i++)
    {
        if((c[i] - '0') == 1)
            b += '0';
        else
            b += '1';
    }
       
    // output
    Console.Write(Convert.ToInt32(b, 2));
     
}
   
// Driver Code
public static void Main(String []args)
{
    int a = 4;
       
    // Function Call
    maxValue(a);
}
}
 
// This code is contributed by 29AjayKumar

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Output: 

3


 

Performance Analysis: 

  • Time Complexity: In the above-given approach, there is conversion from decimal to binary which takes O(logN) time in worst case. Therefore, the time complexity for this approach will be O(logN).
  • Auxiliary Space Complexity: In the above-given approach, there is no extra space used. Therefore, the auxiliary space complexity for the above approach will be O(1)

 

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