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

• Difficulty Level : Easy
• Last Updated : 21 May, 2021

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 are equal to 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 these integers are –
4 ^ 3 = 4 + 3 = 7
4 ^ 2 = 4 + 2 = 6
4 ^ 1 = 4 + 1 = 5
4 ^ 0 = 4 + 0 = 4
The 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 and to get the value of , 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++

 `// C++ implementation to find``// maximum value of B such that``// A ^ B = A + B``#include ``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`

## Java

 `// 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`

## Python3

 `# 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)`

## C#

 `// 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`

## Javascript

 ``
Output:
`3`

Performance Analysis:

• Time Complexity: In the above-given approach, there is the conversion from decimal to binary which takes O(logN) time in the 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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