# Generate n-bit Gray Codes | Set 2

Given a number n, generate bit patterns from 0 to 2^n-1 such that successive patterns differ by one bit.

Examples:

```Input: n=2
Output: 00 01 11 10
Every adjacent element of gray code differs
only by one bit.
So the n bit grey codes are: 00 01 11 10

Input: n=3
Output: 000 001 011 010 110 111 101 100
Every adjacent element of gray code differs
only by one bit.
So the n bit gray codes are:
000 001 011 010 110 111 101 100
```

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

Another approach of Generate n-bit Gray Codes has already been discussed.

Approach:
The idea is to get gray code of binary number using XOR and Right shift operation.

1. The first bit(MSB) of the gray code is same as the first bit(MSB) of binary number.
2. The second bit(from left side) of the gray code equals to XOR of first bit(MSB) and second bit(2nd MSB) of the binary number.
3. The third bit(from left side) of the gray code equals to XOR of the second bit(2nd MSB) and third bit(3rd MSB) and so on..

In this way, the gray code can be calculated for the corresponding binary number. So, it can be observed that the ith element can be formed by bitwise XOR of i and floor(i/2) which is equal to the bitwise XOR of i and (i >> 1) i.e., i right-shifted by 1. By performing this the MSB of the binary number is kept intact and all the other bits are performed bitwise XOR with its adjacent higher bit.

## C++

 `// C++ program to generate n-bit ` `// gray codes ` `#include ` `using` `namespace` `std; ` ` `  `// Function to convert decimal to binary ` `void` `decimalToBinaryNumber(``int` `x, ``int` `n) ` `{ ` `    ``int``* binaryNumber = ``new` `int``(x); ` `    ``int` `i = 0; ` `    ``while` `(x > 0) { ` `        ``binaryNumber[i] = x % 2; ` `        ``x = x / 2; ` `        ``i++; ` `    ``} ` ` `  `    ``// leftmost digits are filled with 0 ` `    ``for` `(``int` `j = 0; j < n - i; j++) ` `        ``cout << ``'0'``; ` ` `  `    ``for` `(``int` `j = i - 1; j >= 0; j--) ` `        ``cout << binaryNumber[j]; ` `} ` ` `  `// Function to generate gray code ` `void` `generateGrayarr(``int` `n) ` `{ ` `    ``int` `N = 1 << n; ` `    ``for` `(``int` `i = 0; i < N; i++) { ` ` `  `        ``// generate gray code of corresponding ` `        ``// binary number of integer i. ` `        ``int` `x = i ^ (i >> 1); ` ` `  `        ``// printing gray code ` `        ``decimalToBinaryNumber(x, n); ` ` `  `        ``cout << endl; ` `    ``} ` `} ` ` `  `// Drivers code ` `int` `main() ` `{ ` `    ``int` `n = 3; ` `    ``generateGrayarr(n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to generate ` `// n-bit gray codes ` `import` `java.io.*; ` ` `  `class` `GFG { ` ` `  `    ``// Function to convert ` `    ``// decimal to binary ` `    ``static` `void` `decimalToBinaryNumber(``int` `x, ` `                                      ``int` `n) ` `    ``{ ` `        ``int``[] binaryNumber = ``new` `int``[x]; ` `        ``int` `i = ``0``; ` `        ``while` `(x > ``0``) { ` `            ``binaryNumber[i] = x % ``2``; ` `            ``x = x / ``2``; ` `            ``i++; ` `        ``} ` ` `  `        ``// leftmost digits are ` `        ``// filled with 0 ` `        ``for` `(``int` `j = ``0``; j < n - i; j++) ` `            ``System.out.print(``'0'``); ` ` `  `        ``for` `(``int` `j = i - ``1``; ` `             ``j >= ``0``; j--) ` `            ``System.out.print(binaryNumber[j]); ` `    ``} ` ` `  `    ``// Function to generate ` `    ``// gray code ` `    ``static` `void` `generateGrayarr(``int` `n) ` `    ``{ ` `        ``int` `N = ``1` `<< n; ` `        ``for` `(``int` `i = ``0``; i < N; i++) { ` ` `  `            ``// generate gray code of ` `            ``// corresponding binary ` `            ``// number of integer i. ` `            ``int` `x = i ^ (i >> ``1``); ` ` `  `            ``// printing gray code ` `            ``decimalToBinaryNumber(x, n); ` ` `  `            ``System.out.println(); ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``3``; ` `        ``generateGrayarr(n); ` `    ``} ` `} ` ` `  `// This code is contributed ` `// by anuj_67. `

## Python3

 `# Python program to generate ` `# n-bit gray codes ` ` `  `# Function to convert ` `# decimal to binary ` `def` `decimalToBinaryNumber(x, n): ` `    ``binaryNumber ``=` `[``0``]``*``x; ` `    ``i ``=` `0``; ` `    ``while` `(x > ``0``): ` `        ``binaryNumber[i] ``=` `x ``%` `2``; ` `        ``x ``=` `x ``/``/` `2``; ` `        ``i ``+``=` `1``; ` ` `  `    ``# leftmost digits are ` `    ``# filled with 0 ` `    ``for` `j ``in` `range``(``0``, n ``-` `i): ` `        ``print``(``'0'``, end ``=``""); ` ` `  `    ``for` `j ``in` `range``(i ``-` `1``, ``-``1``, ``-``1``): ` `        ``print``(binaryNumber[j], end ``=``""); ` ` `  `# Function to generate ` `# gray code ` `def` `generateGrayarr(n): ` `    ``N ``=` `1` `<< n; ` `    ``for` `i ``in` `range``(N): ` `         `  `        ``# generate gray code of ` `        ``# corresponding binary ` `        ``# number of integer i. ` `        ``x ``=` `i ^ (i >> ``1``); ` ` `  `        ``# printing gray code ` `        ``decimalToBinaryNumber(x, n); ` ` `  `        ``print``(); ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n ``=` `3``; ` `    ``generateGrayarr(n); ` ` `  `# This code is contributed by 29AjayKumar `

## C#

 `// C# program to generate ` `// n-bit gray codes ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Function to convert ` `    ``// decimal to binary ` `    ``static` `void` `decimalToBinaryNumber(``int` `x, ` `                                      ``int` `n) ` `    ``{ ` `        ``int``[] binaryNumber = ``new` `int``[x]; ` `        ``int` `i = 0; ` `        ``while` `(x > 0) { ` `            ``binaryNumber[i] = x % 2; ` `            ``x = x / 2; ` `            ``i++; ` `        ``} ` ` `  `        ``// leftmost digits are ` `        ``// filled with 0 ` `        ``for` `(``int` `j = 0; j < n - i; j++) ` `            ``Console.Write(``'0'``); ` ` `  `        ``for` `(``int` `j = i - 1; ` `             ``j >= 0; j--) ` `            ``Console.Write(binaryNumber[j]); ` `    ``} ` ` `  `    ``// Function to generate ` `    ``// gray code ` `    ``static` `void` `generateGrayarr(``int` `n) ` `    ``{ ` `        ``int` `N = 1 << n; ` `        ``for` `(``int` `i = 0; i < N; i++) { ` ` `  `            ``// Generate gray code of ` `            ``// corresponding binary ` `            ``// number of integer i. ` `            ``int` `x = i ^ (i >> 1); ` ` `  `            ``// printing gray code ` `            ``decimalToBinaryNumber(x, n); ` ` `  `            ``Console.WriteLine(); ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 3; ` `        ``generateGrayarr(n); ` `    ``} ` `} ` ` `  `// This code is contributed ` `// by anuj_67. `

Output:

```000
001
011
010
110
111
101
100
```

Complexity Anslysis:

• Time Complexity: O(n).
Only one traversal from 0 to (n-1) is needed.
• Auxiliary Space: O(log x).
A space of (log x) is required for binary representation of (x).

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

9

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.