# 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
```

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

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; ` `} `

## 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 `

Output:

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

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Improved By : SHUBHAMSINGH10