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Generate n-bit Gray Codes
  • Difficulty Level : Medium
  • Last Updated : 28 Dec, 2020

Given a number N, generate bit patterns from 0 to 2^N-1 such that successive patterns differ by one bit. 

Examples:

Input: N = 2
Output: 00 01 11 10

Input: N = 3
Output: 000 001 011 010 110 111 101 100
 

Method-1 

The above sequences are Gray Codes of different widths. Following is an interesting pattern in Gray Codes. 
n-bit Gray Codes can be generated from list of (n-1)-bit Gray codes using following steps. 
1) Let the list of (n-1)-bit Gray codes be L1. Create another list L2 which is reverse of L1. 
2) Modify the list L1 by prefixing a ‘0’ in all codes of L1. 
3) Modify the list L2 by prefixing a ‘1’ in all codes of L2. 
4) Concatenate L1 and L2. The concatenated list is required list of n-bit Gray codes.
For example, following are steps for generating the 3-bit Gray code list from the list of 2-bit Gray code list. 
L1 = {00, 01, 11, 10} (List of 2-bit Gray Codes) 
L2 = {10, 11, 01, 00} (Reverse of L1) 
Prefix all entries of L1 with ‘0’, L1 becomes {000, 001, 011, 010} 
Prefix all entries of L2 with ‘1’, L2 becomes {110, 111, 101, 100} 
Concatenate L1 and L2, we get {000, 001, 011, 010, 110, 111, 101, 100}
To generate n-bit Gray codes, we start from list of 1 bit Gray codes. The list of 1 bit Gray code is {0, 1}. We repeat above steps to generate 2 bit Gray codes from 1 bit Gray codes, then 3-bit Gray codes from 2-bit Gray codes till the number of bits becomes equal to n. 

Below is the implementation of the above approach:



C++

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// C++ program to generate n-bit Gray codes
#include <iostream>
#include <string>
#include <vector>
using namespace std;
 
// This function generates all n bit Gray codes and prints the
// generated codes
void generateGrayarr(int n)
{
    // base case
    if (n <= 0)
        return;
 
    // 'arr' will store all generated codes
    vector<string> arr;
 
    // start with one-bit pattern
    arr.push_back("0");
    arr.push_back("1");
 
    // Every iteration of this loop generates 2*i codes from previously
    // generated i codes.
    int i, j;
    for (i = 2; i < (1<<n); i = i<<1)
    {
        // Enter the prviously generated codes again in arr[] in reverse
        // order. Nor arr[] has double number of codes.
        for (j = i-1 ; j >= 0 ; j--)
            arr.push_back(arr[j]);
 
        // append 0 to the first half
        for (j = 0 ; j < i ; j++)
            arr[j] = "0" + arr[j];
 
        // append 1 to the second half
        for (j = i ; j < 2*i ; j++)
            arr[j] = "1" + arr[j];
    }
 
    // print contents of arr[]
    for (i = 0 ; i < arr.size() ; i++ )
        cout << arr[i] << endl;
}
 
// Driver program to test above function
int main()
{
    generateGrayarr(3);
    return 0;
}

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Java

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// Java program to generate n-bit Gray codes
import java.util.*;
class GfG {
 
// This function generates all n bit Gray codes and prints the
// generated codes
static void generateGrayarr(int n)
{
    // base case
    if (n <= 0)
        return;
 
    // 'arr' will store all generated codes
    ArrayList<String> arr = new ArrayList<String> ();
 
    // start with one-bit pattern
    arr.add("0");
    arr.add("1");
 
    // Every iteration of this loop generates 2*i codes from previously
    // generated i codes.
    int i, j;
    for (i = 2; i < (1<<n); i = i<<1)
    {
        // Enter the prviously generated codes again in arr[] in reverse
        // order. Nor arr[] has double number of codes.
        for (j = i-1 ; j >= 0 ; j--)
            arr.add(arr.get(j));
 
        // append 0 to the first half
        for (j = 0 ; j < i ; j++)
            arr.set(j, "0" + arr.get(j));
 
        // append 1 to the second half
        for (j = i ; j < 2*i ; j++)
            arr.set(j, "1" + arr.get(j));
    }
 
    // print contents of arr[]
    for (i = 0 ; i < arr.size() ; i++ )
        System.out.println(arr.get(i));
}
 
// Driver program to test above function
public static void main(String[] args)
{
    generateGrayarr(3);
}
}

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Python3

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# Python3 program to generate n-bit Gray codes
import math as mt
 
# This function generates all n bit Gray
# codes and prints the generated codes
def generateGrayarr(n):
 
    # base case
    if (n <= 0):
        return
 
    # 'arr' will store all generated codes
    arr = list()
 
    # start with one-bit pattern
    arr.append("0")
    arr.append("1")
 
    # Every iteration of this loop generates
    # 2*i codes from previously generated i codes.
    i = 2
    j = 0
    while(True):
 
        if i >= 1 << n:
            break
     
        # Enter the prviously generated codes
        # again in arr[] in reverse order.
        # Nor arr[] has double number of codes.
        for j in range(i - 1, -1, -1):
            arr.append(arr[j])
 
        # append 0 to the first half
        for j in range(i):
            arr[j] = "0" + arr[j]
 
        # append 1 to the second half
        for j in range(i, 2 * i):
            arr[j] = "1" + arr[j]
        i = i << 1
 
    # prcontents of arr[]
    for i in range(len(arr)):
        print(arr[i])
 
# Driver Code
generateGrayarr(3)
 
# This code is contributed
# by Mohit kumar 29

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C#

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using System;
using System.Collections.Generic;
 
// C# program to generate n-bit Gray codes 
public class GfG
{
 
// This function generates all n bit Gray codes and prints the 
// generated codes 
public static void generateGrayarr(int n)
{
    // base case 
    if (n <= 0)
    {
        return;
    }
 
    // 'arr' will store all generated codes 
    List<string> arr = new List<string> ();
 
    // start with one-bit pattern 
    arr.Add("0");
    arr.Add("1");
 
    // Every iteration of this loop generates 2*i codes from previously 
    // generated i codes. 
    int i, j;
    for (i = 2; i < (1 << n); i = i << 1)
    {
        // Enter the prviously generated codes again in arr[] in reverse 
        // order. Nor arr[] has double number of codes. 
        for (j = i - 1 ; j >= 0 ; j--)
        {
            arr.Add(arr[j]);
        }
 
        // append 0 to the first half 
        for (j = 0 ; j < i ; j++)
        {
            arr[j] = "0" + arr[j];
        }
 
        // append 1 to the second half 
        for (j = i ; j < 2 * i ; j++)
        {
            arr[j] = "1" + arr[j];
        }
    }
 
    // print contents of arr[] 
    for (i = 0 ; i < arr.Count ; i++)
    {
        Console.WriteLine(arr[i]);
    }
}
 
// Driver program to test above function 
public static void Main(string[] args)
{
    generateGrayarr(3);
}
}
 
  // This code is contributed by Shrikant13

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Output

000
001
011
010
110
111
101
100

Method 2: Recursive Approach

The idea is to recursively append the bit 0 and 1 each time until the number of bits is not equal to N. 

Base Condition: The base case for this problem will be when the value of N = 0 or 1. 

If (N == 0)
       return {“0”}
if (N == 1)
     return {“0”, “1”}

Recursive Condition: Otherwise, for any value greater than 1, recursively generate the gray codes of the N – 1 bits and then for each of the gray code generated add the prefix 0 and 1.

Below is the implementation of the above approach: 

C++

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// C++ program to generate
// n-bit Gray codes
 
#include <bits/stdc++.h>
using namespace std;
 
// This function generates all n
// bit Gray codes and prints the
// generated codes
vector<string> generateGray(int n)
{
    // Base case
    if (n <= 0)
        return {"0"};
 
    if (n == 1)
    {
      return {"0","1"};
    }
 
    //Recursive case
    vector<string> recAns=
          generateGray(n-1);
    vector<string> mainAns;
     
    // Append 0 to the first half
    for(int i=0;i<recAns.size();i++)
    {
      string s=recAns[i];
      mainAns.push_back("0"+s);
    }
     
     // Append 1 to the second half
    for(int i=recAns.size()-1;i>=0;i--)
    {
       string s=recAns[i];
       mainAns.push_back("1"+s);
    }
    return mainAns;
}
 
// Function to generate the
// Gray code of N bits
void generateGrayarr(int n)
{
    vector<string> arr;
    arr=generateGray(n);
    // print contents of arr
    for (int i = 0 ; i < arr.size();
         i++ )
        cout << arr[i] << endl;
}
 
// Driver Code
int main()
{
    generateGrayarr(3);
    return 0;
}

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Python3

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# Python3 program to generate
# n-bit Gray codes
 
# This function generates all n
# bit Gray codes and prints the
# generated codes
def generateGray(n):
     
    # Base case
    if (n <= 0):
        return ["0"]
    if (n == 1):
        return [ "0", "1" ]
 
    # Recursive case
    recAns = generateGray(n - 1)
 
    mainAns = []
     
    # Append 0 to the first half
    for i in range(len(recAns)):
        s = recAns[i]
        mainAns.append("0" + s)
 
    # Append 1 to the second half
    for i in range(len(recAns) - 1, -1, -1):
        s = recAns[i]
        mainAns.append("1" + s)
 
    return mainAns
 
# Function to generate the
# Gray code of N bits
def generateGrayarr(n):
     
    arr = generateGray(n)
 
    # Print contents of arr
    print(*arr, sep = "\n")
 
# Driver Code
generateGrayarr(3)
 
# This code is contributed by avanitrachhadiya2155

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Output



000
001
011
010
110
111
101
100

Method3: (Using bitset)

We should first find binary no from 1 to n and then convert it into string and then print it using substring function of string.

Below is the implementation of the above idea:

C++

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#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    n = 5;
   
    // power of 2
    int k = (log2(n)) + 1;
    vector<string> res;
    for (int i = 0; i <= n; i++)
    {
        // using bitset
        std::bitset<32> r(i);
       
        // converting to string
        string s = r.to_string();
        cout << s.substr(32 - k) << " ";
    }
 
    return 0;
}

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Output

000 001 010 011 100 101

Time Complexity: O(n)

This article is compiled by Ravi Chandra Enaganti. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

 

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