# A backtracking approach to generate n bit Gray Codes

• Difficulty Level : Hard
• Last Updated : 15 Jun, 2022

Given a number n, the task is to generate n bit Gray codes (generate bit patterns from 0 to 2^n-1 such that successive patterns differ by one bit)

Examples:

```Input : 2
Output : 0 1 3 2
Explanation :
00 - 0
01 - 1
11 - 3
10 - 2

Input : 3
Output : 0 1 3 2 6 7 5 4
```

We have discussed an approach in Generate n-bit Gray Codes
This article provides a backtracking approach to the same problem. Idea is that for each bit out of n bit we have a choice either we can ignore it or we can invert the bit so this means our gray sequence goes upto 2 ^ n for n bits. So we make two recursive calls for either inverting the bit or leaving the bit as it is.

## C++

 `// CPP program to find the gray sequence of n bits.``#include ``#include ``using` `namespace` `std;` `/* we have 2 choices for each of the n bits either we``   ``can include i.e invert the bit or we can exclude the``   ``bit i.e we can leave the number as it is. */``void` `grayCodeUtil(vector<``int``>& res, ``int` `n, ``int``& num)``{``    ``// base case when we run out bits to process``    ``// we simply include it in gray code sequence.``    ``if` `(n == 0) {``        ``res.push_back(num);``        ``return``;``    ``}` `    ``// ignore the bit.``    ``grayCodeUtil(res, n - 1, num);` `    ``// invert the bit.``    ``num = num ^ (1 << (n - 1));``    ``grayCodeUtil(res, n - 1, num);``}` `// returns the vector containing the gray``// code sequence of n bits.``vector<``int``> grayCodes(``int` `n)``{``    ``vector<``int``> res;` `    ``// num is passed by reference to keep``    ``// track of current code.``    ``int` `num = 0;``    ``grayCodeUtil(res, n, num);` `    ``return` `res;``}` `// Driver function.``int` `main()``{``    ``int` `n = 3;``    ``vector<``int``> code = grayCodes(n);``    ``for` `(``int` `i = 0; i < code.size(); i++)``        ``cout << code[i] << endl;   ``    ``return` `0;``}`

## Java

 `// JAVA program to find the gray sequence of n bits.``import` `java.util.*;` `class` `GFG``{` `static` `int` `num;` `/* we have 2 choices for each of the n bits either we``can include i.e invert the bit or we can exclude the``bit i.e we can leave the number as it is. */``static` `void` `grayCodeUtil(Vector res, ``int` `n)``{``    ``// base case when we run out bits to process``    ``// we simply include it in gray code sequence.``    ``if` `(n == ``0``)``    ``{``        ``res.add(num);``        ``return``;``    ``}` `    ``// ignore the bit.``    ``grayCodeUtil(res, n - ``1``);` `    ``// invert the bit.``    ``num = num ^ (``1` `<< (n - ``1``));``    ``grayCodeUtil(res, n - ``1``);``}` `// returns the vector containing the gray``// code sequence of n bits.``static` `Vector grayCodes(``int` `n)``{``    ``Vector res = ``new` `Vector();` `    ``// num is passed by reference to keep``    ``// track of current code.``    ``num = ``0``;``    ``grayCodeUtil(res, n);` `    ``return` `res;``}` `// Driver function.``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``3``;``    ``Vector code = grayCodes(n);``    ``for` `(``int` `i = ``0``; i < code.size(); i++)``        ``System.out.print(code.get(i) +``"\n"``);``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 program to find the``# gray sequence of n bits.` `""" we have 2 choices for each of the n bits``either we can include i.e invert the bit or``we can exclude the bit i.e we can leave``the number as it is. """``def` `grayCodeUtil(res, n, num):``    ` `    ``# base case when we run out bits to process``    ``# we simply include it in gray code sequence.``    ``if` `(n ``=``=` `0``):``        ``res.append(num[``0``])``        ``return``        ` `    ``# ignore the bit.``    ``grayCodeUtil(res, n ``-` `1``, num)``    ` `    ``# invert the bit.``    ``num[``0``] ``=` `num[``0``] ^ (``1` `<< (n ``-` `1``))``    ``grayCodeUtil(res, n ``-` `1``, num)``    ` `# returns the vector containing the gray``# code sequence of n bits.``def` `grayCodes(n):``    ``res ``=` `[]``    ` `    ``# num is passed by reference to keep``    ``# track of current code.``    ``num ``=` `[``0``]``    ``grayCodeUtil(res, n, num)``    ``return` `res` `# Driver Code``n ``=` `3``code ``=` `grayCodes(n)``for` `i ``in` `range``(``len``(code)):``    ``print``(code[i])` `# This code is contributed by SHUBHAMSINGH10`

## C#

 `// C# program to find the gray sequence of n bits.``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `static` `int` `num;` `/* we have 2 choices for each of the n bits either we``can include i.e invert the bit or we can exclude the``bit i.e we can leave the number as it is. */``static` `void` `grayCodeUtil(List<``int``> res, ``int` `n)``{``    ``// base case when we run out bits to process``    ``// we simply include it in gray code sequence.``    ``if` `(n == 0)``    ``{``        ``res.Add(num);``        ``return``;``    ``}` `    ``// ignore the bit.``    ``grayCodeUtil(res, n - 1);` `    ``// invert the bit.``    ``num = num ^ (1 << (n - 1));``    ``grayCodeUtil(res, n - 1);``}` `// returns the vector containing the gray``// code sequence of n bits.``static` `List<``int``> grayCodes(``int` `n)``{``    ``List<``int``> res = ``new` `List<``int``>();` `    ``// num is passed by reference to keep``    ``// track of current code.``    ``num = 0;``    ``grayCodeUtil(res, n);` `    ``return` `res;``}` `// Driver function.``public` `static` `void` `Main(String[] args)``{``    ``int` `n = 3;``    ``List<``int``> code = grayCodes(n);``    ``for` `(``int` `i = 0; i < code.Count; i++)``        ``Console.Write(code[i] +``"\n"``);``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

```0
1
3
2
6
7
5
4```

Time Complexity: O(n)
Auxiliary Space: O(n)

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