Given a number n, convert it from binary to Gray code and also convert n into Gray code to binary.
Input : n = 5
Output: Binary to Gray : 7
Gray to Binary : 6
Explanation: Binary to Gray gives 7 (binary 101 → gray 111), and gray to binary converts gray code 5 (binary 101) to binary 6 (110)
Try It Yourself
Binary to Gray conversion :
- The Most Significant Bit (MSB) of the gray code is always equal to the MSB of the given binary code.
- Other bits of the output gray code can be obtained by XORing binary code bit at that index and previous index.

Gray to binary conversion :
- The Most Significant Bit (MSB) of the binary code is always equal to the MSB of the given gray code.
- Other bits of the output binary code can be obtained by checking the gray code bit at that index. If the current gray code bit is 0, then copy the previous binary code bit, else copy the invert of the previous binary code bit.

Table of Content
When n is given in binary
#include <iostream>
using namespace std;
// Function to find xor of two
// bits represented as character.
char xorChar(char a, char b) {
return (a == b) ? '0' : '1';
}
// Function to flip a bit
// represented as character.
char flip(char c) {
return (c == '0') ? '1' : '0';
}
// function to convert binary string
// to gray string
string binToGrey(string binary) {
string gray = "";
// MSB of gray code is same as binary code
gray += binary[0];
// Compute remaining bits, next bit is computed by
// doing XOR of previous and current in Binary
for (int i = 1; i < binary.length(); i++) {
// Concatenate XOR of previous bit
// with current bit
gray += xorChar(binary[i - 1], binary[i]);
}
return gray;
}
// function to convert gray code string
// to binary string
string greyToBin(string gray) {
string binary = "";
// MSB of binary code is same as gray code
binary += gray[0];
// Compute remaining bits
for (int i = 1; i < gray.length(); i++) {
// If current bit is 0, concatenate
// previous bit
if (gray[i] == '0')
binary += binary[i - 1];
// Else, concatenate invert of
// previous bit
else
binary += flip(binary[i - 1]);
}
return binary;
}
int main() {
string binary = "01001";
cout <<"Binary to Gray: " << binToGrey(binary) << endl;
string gray = "01101";
cout <<"Gray to Binary: " <<greyToBin(gray) << endl;
return 0;
}
class GfG {
// Function to find xor of two
// bits represented as character.
static char xorChar(char a, char b) {
return (a == b) ? '0' : '1';
}
// Function to flip a bit
// represented as character.
static char flip(char c) {
return (c == '0') ? '1' : '0';
}
// function to convert binary string
// to gray string
static String binToGrey(String binary) {
String gray = "";
// MSB of gray code is same as binary code
gray += binary.charAt(0);
// Compute remaining bits, next bit is computed by
// doing XOR of previous and current in Binary
for (int i = 1; i < binary.length(); i++) {
// Concatenate XOR of previous bit
// with current bit
gray += xorChar(binary.charAt(i - 1), binary.charAt(i));
}
return gray;
}
// function to convert gray code string
// to binary string
static String greyToBin(String gray) {
String binary = "";
// MSB of binary code is same as gray code
binary += gray.charAt(0);
// Compute remaining bits
for (int i = 1; i < gray.length(); i++) {
// If current bit is 0, concatenate
// previous bit
if (gray.charAt(i) == '0')
binary += binary.charAt(i - 1);
// Else, concatenate invert of
// previous bit
else
binary += flip(binary.charAt(i - 1));
}
return binary;
}
public static void main(String[] args) {
String binary = "01001";
System.out.println("Binary to Gray: " + binToGrey(binary));
String gray = "01101";
System.out.println("Gray to Binary: " + greyToBin(gray));
}
}
# Function to find xor of two
# bits represented as character.
def xorChar(a, b):
return '0' if a == b else '1'
# Function to flip a bit
# represented as character.
def flip(c):
return '1' if c == '0' else '0'
# function to convert binary string
# to gray string
def binToGrey(binary):
gray = ""
# MSB of gray code is same as binary code
gray += binary[0]
# Compute remaining bits, next bit is computed by
# doing XOR of previous and current in Binary
for i in range(1, len(binary)):
# Concatenate XOR of previous bit
# with current bit
gray += xorChar(binary[i - 1], binary[i])
return gray
# function to convert gray code string
# to binary string
def greyToBin(gray):
binary = ""
# MSB of binary code is same as gray code
binary += gray[0]
# Compute remaining bits
for i in range(1, len(gray)):
# If current bit is 0, concatenate
# previous bit
if gray[i] == '0':
binary += binary[i - 1]
# Else, concatenate invert of
# previous bit
else:
binary += flip(binary[i - 1])
return binary
if __name__ == "__main__":
binary = "01001"
print("Binary to Gray: " +binToGrey(binary))
gray = "01101"
print("Gray to Binary: " + greyToBin(gray))
using System;
using System.Text;
class GfG {
// Function to find XOR of two
// bits represented as characters
static char xorChar(char a, char b) {
return (a == b) ? '0' : '1';
}
// Function to flip a bit
// represented as a character
static char flip(char c) {
return (c == '0') ? '1' : '0';
}
// Function to convert binary
// string to gray string
static string binToGrey(string binary) {
StringBuilder gray = new StringBuilder();
// MSB of gray code is
// same as binary code
gray.Append(binary[0]);
// Compute remaining bits
for (int i = 1; i < binary.Length; i++) {
gray.Append(xorChar(binary[i - 1], binary[i]));
}
return gray.ToString();
}
// Function to convert gray code
// string to binary string
static string greyToBin(string gray) {
StringBuilder binary = new StringBuilder();
// MSB of binary code is same as gray code
binary.Append(gray[0]);
// Compute remaining bits
for (int i = 1; i < gray.Length; i++) {
if (gray[i] == '0')
binary.Append(binary[i - 1]);
else
binary.Append(flip(binary[i - 1]));
}
return binary.ToString();
}
static void Main() {
string binary = "01001";
Console.WriteLine("Binary to Gray: " + binToGrey(binary));
string gray = "01101";
Console.WriteLine("Gray to Binary: " + greyToBin(gray));
}
}
// JavaScript program for Binary To Gray
// and Gray to Binary conversion
// Function to find xor of two
// bits represented as character.
function xorChar(a, b) {
return (a === b) ? '0' : '1';
}
// Function to flip a bit
// represented as character.
function flip(c) {
return (c === '0') ? '1' : '0';
}
// function to convert binary string
// to gray string
function binToGrey(binary) {
let gray = "";
// MSB of gray code is same as binary code
gray += binary[0];
// Compute remaining bits, next bit is computed by
// doing XOR of previous and current in Binary
for (let i = 1; i < binary.length; i++) {
// Concatenate XOR of previous bit
// with current bit
gray += xorChar(binary[i - 1], binary[i]);
}
return gray;
}
// function to convert gray code string
// to binary string
function greyToBin(gray) {
let binary = "";
// MSB of binary code is same as gray code
binary += gray[0];
// Compute remaining bits
for (let i = 1; i < gray.length; i++) {
// If current bit is 0, concatenate
// previous bit
if (gray[i] === '0')
binary += binary[i - 1];
// Else, concatenate invert of
// previous bit
else
binary += flip(binary[i - 1]);
}
return binary;
}
// Driver Code
let binary = "01001";
console.log("Binary to Gray: "+ binToGrey(binary));
let gray = "01101";
console.log("Gray to Binary: " + greyToBin(gray));
Output
Binary to Gray: 01101 Gray to Binary: 01001
Time Complexity: O(n), where n is length of the binary string.
Auxiliary Space: O(n)
When n is given in interger
#include <iostream>
using namespace std;
int binToGrey(int n) {
return n ^ (n >> 1);
}
int greyToBin(int n) {
int res = n;
while (n > 0) {
n >>= 1;
res ^= n;
}
return res;
}
int main() {
int n = 5;
cout << "Binary to Gray: " <<binToGrey(n) << endl;
cout << "Gray to Binary: " <<greyToBin(n) << endl;
return 0;
}
// Java program for Binary To Gray
// and Gray to Binary conversion
class GfG {
// Function to convert binary to gray
static int binToGrey(int n) {
return n ^ (n >> 1);
}
// Function to convert gray to binary
static int greyToBin(int n) {
int res = n;
while (n > 0) {
n >>= 1;
res ^= n;
}
return res;
}
public static void main(String[] args) {
int n = 5;
System.out.println("Binary to Gray: " + binToGrey(n));
System.out.println("Gray to Binary: " + greyToBin(n));
}
}
def binToGrey(n):
return n ^ (n >> 1)
# Function to convert gray to binary
def greyToBin(n):
res = n
while n > 0:
n >>= 1
res ^= n
return res
if __name__ == "__main__":
n = 5
print("Binary to Gray:",binToGrey(n))
print("Gray to Binary:",greyToBin(n))
// C# program for Binary To Gray
// and Gray to Binary conversion
using System;
class GfG {
// Function to convert binary to gray
static int binToGrey(int n) {
return n ^ (n >> 1);
}
// Function to convert gray to binary
static int greyToBin(int n) {
int res = n;
while (n > 0) {
n >>= 1;
res ^= n;
}
return res;
}
static void Main() {
int n = 5;
Console.WriteLine("Binary to Gray: " + binToGrey(n));
Console.WriteLine("Gray to Binary: "+ greyToBin(n));
}
}
// JavaScript program for Binary To Gray
// and Gray to Binary conversion
// Function to convert binary to gray
function binToGrey(n) {
return n ^ (n >> 1);
}
// Function to convert gray to binary
function greyToBin(n) {
let res = n;
while (n > 0) {
n >>= 1;
res ^= n;
}
return res;
}
// Driver Code
let binary = 5;
console.log("Binary to Gray: " + binToGrey(n));
console.log("Gray to Binary: " + greyToBin(n));
Output
Binary to Gray: 7 Gray to Binary: 6
Time Complexity:
- For binary to gray conversion: O(1)
- For gray to binary conversion: O(log(n)), as n is a decimal number and number of bits is approximately log(n).
Auxiliary Space: O(1)
Related Article: