How to print n-bit gray codes specifically using backtracking?

Question

I codded a program to print n-bit binary gray codes. But I'm not sure if it is a backtracking program. If it is not a backtracking program, what is something I can do to use it in this code.

I want to print the binary codes themselves, not their decimal equivalent.

                vector<string> grayCode(int n)
                {
                
                    if(n<=0)
                        return {"0"};
                    if(n == 1)
                        return {"0", "1"};
                    
                    vector<string> list1 = grayCode(n-1);
                    
                    vector<string> mainList;
                    
                    for(int i=0; i<list1.size(); i++)
                        mainList.push_back("0" + list1[i]);
                        
                    for(int i=list1.size()-1; i>-1; i--){
                        mainList.push_back("1" + list1[i]);
                    
                    return mainList;
                }
                
                int main()
                {
                    vector<string> gcode = grayCode(4);
                    for(int i=0; i<gcode.size(); i++)
                        cout<< gcode[i] << "    ";
                    
                    return 0;
                }

Answer 1

I want to print the binary codes themselves, not their decimal equivalent.

That should be your question. None of it has to do with grey code or backtracking. It's a simple output formatting question.

You can change the output format using for example std::hex like this:

std::cout << std::hex << gcode[i] << "    ";

That will output the value in hexadecimal. There is also std::oct for octal. Sadly there is no std::bin for binary. And std::cout.setbase(2); will also not work.

I think the simplest is to use std::bitset, specifically the to_string().