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# Binary to Gray code using recursion

Given the Binary code of a number as a decimal number, we need to convert this into its equivalent Gray Code. Assume that the binary number is in the range of integers. For the larger value, we can take a binary number as string.

In gray code, only one bit is changed in 2 consecutive numbers.

Examples:

Input: 1001
Output: 1101
Explanation: 1001 -> 1101 -> 1101 -> 1101

Input: 11
Output: 10
Explanation: 11 -> 10

## Approach:

The idea is to check whether the last bit and second last bit are same or not, if it is same then move ahead otherwise add 1.

Follow the steps to solve the given problem:

• binary_to_grey(n)
•  if n == 0
•  grey = 0;
•  else if last two bits are opposite  to each other
•  grey = 1 + 10 * binary_to_gray(n/10))
•  else if last two bits are same
•   grey = 10 * binary_to_gray(n/10))

Below is the implementation of the above approach :

## CPP

 `// CPP program to convert Binary to``// Gray code using recursion``#include ``using` `namespace` `std;` `// Function to change Binary to``// Gray using recursion``int` `binary_to_gray(``int` `n)``{``    ``if` `(!n)``        ``return` `0;` `    ``// Taking last digit``    ``int` `a = n % 10;` `    ``// Taking second last digit``    ``int` `b = (n / 10) % 10;` `    ``// If last digit are opposite bits``    ``if` `((a && !b) || (!a && b))``        ``return` `(1 + 10 * binary_to_gray(n / 10));` `    ``// If last two bits are same``    ``return` `(10 * binary_to_gray(n / 10));``}` `// Driver Function``int` `main()``{``    ``int` `binary_number = 1011101;` `    ``printf``(``"%d"``, binary_to_gray(binary_number));``    ``return` `0;``}`

## Java

 `// Java program to convert``// Binary code to Gray code``import` `static` `java.lang.StrictMath.pow;` `import` `java.util.Scanner;` `class` `bin_gray {``    ``// Function to change Binary to``    ``// Gray using recursion``    ``int` `binary_to_gray(``int` `n, ``int` `i)``    ``{``        ``int` `a, b;``        ``int` `result = ``0``;``        ``if` `(n != ``0``) {``            ``// Taking last digit``            ``a = n % ``10``;` `            ``n = n / ``10``;` `            ``// Taking second last digit``            ``b = n % ``10``;` `            ``if` `((a & ~b) == ``1` `|| (~a & b) == ``1``) {``                ``result = (``int``)(result + pow(``10``, i));``            ``}` `            ``return` `binary_to_gray(n, ++i) + result;``        ``}``        ``return` `0``;``    ``}` `    ``// Driver Function``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `binary_number;``        ``int` `result = ``0``;``        ``binary_number = ``1011101``;` `        ``bin_gray obj = ``new` `bin_gray();``        ``result = obj.binary_to_gray(binary_number, ``0``);` `        ``System.out.print(result);``    ``}``}` `// This article is contributed by Anshika Goyal.`

## Python3

 `# Python3 code to convert Binary``# to Gray code using recursion` `# Function to change Binary to Gray using recursion`  `def` `binary_to_gray(n):``    ``if` `not``(n):``        ``return` `0` `    ``# Taking last digit``    ``a ``=` `n ``%` `10` `    ``# Taking second last digit``    ``b ``=` `int``(n ``/` `10``) ``%` `10` `    ``# If last digit are opposite bits``    ``if` `(a ``and` `not``(b)) ``or` `(``not``(a) ``and` `b):``        ``return` `(``1` `+` `10` `*` `binary_to_gray(``int``(n ``/` `10``)))` `    ``# If last two bits are same``    ``return` `(``10` `*` `binary_to_gray(``int``(n ``/` `10``)))`  `# Driver Code``binary_number ``=` `1011101``print``(binary_to_gray(binary_number), end``=``'')` `# This code is contributed by "Sharad_Bhardwaj".`

## C#

 `// C# program to convert``// Binary code to Gray code``using` `System;` `class` `GFG {` `    ``// Function to change Binary to``    ``// Gray using recursion``    ``static` `int` `binary_to_gray(``int` `n, ``int` `i)``    ``{``        ``int` `a, b;``        ``int` `result = 0;``        ``if` `(n != 0) {` `            ``// Taking last digit``            ``a = n % 10;` `            ``n = n / 10;` `            ``// Taking second last digit``            ``b = n % 10;` `            ``if` `((a & ~b) == 1 || (~a & b) == 1) {``                ``result = (``int``)(result + Math.Pow(10, i));``            ``}` `            ``return` `binary_to_gray(n, ++i) + result;``        ``}` `        ``return` `0;``    ``}` `    ``// Driver Function``    ``public` `static` `void` `Main()``    ``{``        ``int` `binary_number;``        ``binary_number = 1011101;` `        ``Console.WriteLine(binary_to_gray(binary_number, 0));``    ``}``}` `// This article is contributed by vt_m.`

## PHP

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## Javascript

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Output

`1110011`

Time Complexity: O(logN), Traverse through all the digits, as there are logN bits.
Auxiliary Space: O(1)