# 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 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`

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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