# 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: 3
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: 7
Output: 0
There is no integer except 0 such that A + B = A ^ B

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

## 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 `

## Python

 `# Python 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).
• Auxillary Space Complexity: In the above-given approach, there is no extra space used. Therefore, the auxillary space complexity for the above approach will be O(1)

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