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# A backtracking approach to generate n bit Gray Codes

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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