Given a number N, write a function which generates all n-bit gray code sequences, a gray code sequence is a sequence such that successive patterns in it differ by one bit

Construction an n bit gray code follows a robust algorithm. The binary-reflected Gray code list for n bits can be generated recursively from the list for n − 1 bits by reflecting the list (i.e. listing the entries in reverse order), prefixing the entries in the original list with a binary 0, prefixing the entries in the reflected list with a binary 1, and then concatenating the original list with the reversed list.

Reference: Gray_code

Algorithm to generate n bit gray code list

1. Let, the n-1 length gray code list has m elements, (m=2^(n-1))
2. Create list1 by adding prefix 0 to all the m elements
3. Create list2 by adding prefix 1 to all the m elements
4. New list for n-bit gray code= list1+reverse(list)
5. New list size 2m (2^n)

For any n bit gray code list, we start with gray code list of length 1 (elements are only '0' and '1'). Then iterate till finding n-bit gray code list.

The above can be explained with help of an example,

    Let we need to find gray code sequence for n=3
    //Base case
    For n=1
    List: 0, 1

    For n=2
    Add prefix '0' to create list1
    List1= 00, 01
    Add prefix '1' to create list2
    List2= 10, 11
    New list=list1 + reverse(list2)
    =00, 01, 11, 10 //List for n=2

    For n=3
    Add prefix '0' to create list1
    List1= 000, 001, 011, 010
    Add prefix '1' to create list2
    List2= 100, 101, 111, 110
    New list=list1 + reverse(list2)
    =000, 001, 011, 010, 110, 111, 101, 100 //List for n=3

C++ implementation

#include <bits/stdc++.h>
using namespace std;

string get_string(char c)
{
	string s(1,c);
	return s;
}

void generateCode(int n)
{
	if(n<=0){
		cout<<"invalid input\n";
		return;
	}

	vector<string> a,temp;
	a.push_back("0");
	a.push_back("1");

	for(int i=0;i<n-1;i++){
		for(int j=0;j<a.size();j++){
			temp.push_back(get_string('0')+a[j]);
		}
		for(int j=a.size()-1;j>=0;j--){
			temp.push_back(get_string('1')+a[j]);
		}
		a=temp;
		temp.clear();
	}

	for(int i=0;i<a.size();i++)
		cout<<a[i]<<"\n";
}

int main(){
	int n;
	
	cout<<"Enter n:\n";
	cin>>n;
	
	cout<<"Printing gray codes for "<<n<<" bit\n";
	generateCode(n);

	return 0;
}

Output

First run:
Enter n:
3
Printing gray codes for 3 bit
000
001
011
010
110
111
101
100


Second run:
Enter n:
4
Printing gray codes for 4 bit
0000
0001
0011
0010
0110
0111
0101
0100
1100
1101
1111
1110
1010
1011
1001
1000