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 <iostream>
#include <vector>
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<Integer> 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<Integer> grayCodes(int n)
{
    Vector<Integer> res = new Vector<Integer>();

    // 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<Integer> 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

<script>

// Javascript 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. */
function 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.push(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.
function grayCodes(n)
{
    let res = [];

    // num is passed by reference to keep
    // track of current code.
    let num = [0];
    grayCodeUtil(res, n, num);

    return res;
}

// Driver code
let n = 3;
let code = grayCodes(n);
for(let i = 0; i < code.length; i++)
    document.write(code[i] + "<br>");

// This code is contributed by gfgking

</script>

Output:

0
1
3
2
6
7
5
4

Time Complexity: O(2ⁿ)
Auxiliary Space: O(n)