Gray to Binary Conversion

Given a non-negative integer n, whose binary representation represents a Gray code, convert this Gray code to its equivalent binary number and return the decimal value of that binary number.

Examples:

Input: n = 4
Output: 7
Explanation: The binary representation of 4 is 100. Treating 100 as a Gray code, its corresponding binary representation is 111, whose decimal value is 7.

Input: n = 7
Output: 5
Explanation: The binary representation of 7 is 111. Treating 111 as a Gray code, its corresponding binary representation is 101, whose decimal value is 5.

The following table shows the conversion of Gray code values to their corresponding binary values:

Using Bit Manipulation - O(log(n)) Time and O(1) Space

The most significant bit of a binary number is the same as the most significant bit of its Gray code. Every subsequent binary bit can be obtained by XORing the previous binary bit with the corresponding Gray code bit.

Based on this property, the original binary number can be reconstructed by taking the cumulative XOR of the Gray code and all of its right-shifted values:

Binary = Gray ^ (Gray >> 1) ^ (Gray >> 2) ^ ...

We initialize a variable b as 0. Then, we repeatedly XOR b with the current value of n and right shift n by one bit. This process continues until n becomes 0. After all bits have been processed, b stores the decimal value of the equivalent binary number.

Consider n = 7.

The binary representation of 7 is 111, which is treated as a Gray code.

• b = 0, b = b ^ 111 = 111
• n = 111 >> 1 = 011, b = b ^ 011 = 100
• n = 011 >> 1 = 001, b = b ^ 001 = 101
• n = 001 >> 1 = 000 stop here because n=0.

The resulting binary representation is 101, whose decimal value is 5.

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

int grayToBinary(int n) {
    int b = 0;

    // Traverse all bits of Gray code
    while (n > 0) {

        // Build binary number step by step using XOR
        b ^= n;

        // Move to next bit
        n = n >> 1;
    }

    return b;
}

int main() {
    int n = 7;

    cout << grayToBinary(n);

    return 0;
}

#include <stdio.h>

int grayToBinary(int n) {
    int b = 0;

    // Traverse all bits of Gray code
    while (n > 0) {

        // Build binary number step by step using XOR
        b ^= n;

        // Move to next bit
        n = n >> 1;
    }

    return b;
}

int main() {
    int n = 7;

    printf("%d", grayToBinary(n));

    return 0;
}

class GFG {

    static int grayToBinary(int n) {
        int b = 0;

        // Traverse all bits of Gray code
        while (n > 0) {

            // Build binary number step by step using XOR
            b ^= n;

            // Move to next bit
            n = n >> 1;
        }

        return b;
    }

    public static void main(String[] args) {
        int n = 7;

        System.out.print(grayToBinary(n));
    }
}

def grayToBinary(n):
    b = 0

    # Traverse all bits of Gray code
    while n > 0:

        # Build binary number step by step using XOR
        b ^= n

        # Move to next bit
        n = n >> 1

    return b


if __name__ == "__main__":
    n = 7

    print(grayToBinary(n))

using System;

class GFG
{
    static int grayToBinary(int n)
    {
        int b = 0;

        // Traverse all bits of Gray code
        while (n > 0)
        {

            // Build binary number step by step using XOR
            b ^= n;

            // Move to next bit
            n = n >> 1;
        }

        return b;
    }

    static void Main()
    {
        int n = 7;

        Console.Write(grayToBinary(n));
    }
}

function grayToBinary(n) {
    let b = 0;

    // Traverse all bits of Gray code
    while (n > 0) {

        // Build binary number step by step using XOR
        b ^= n;

        // Move to next bit
        n = n >> 1;
    }

    return b;
}

// Driver code

    let n = 7;

    console.log(grayToBinary(n));

5