Find the remainder when N is divided by 4 using Bitwise AND operator

Given a number N, the task is to find the remainder when N is divided by 4 using Bitwise AND operator.

Examples:

Input: N = 98 
Output: 2
Explanation: 98 % 4 = 2. Hence the output is 2.

Input: 200
Output: 0
Explanation: 200 % 4 = 0. Hence output is 0.

Naive approach:
For solving the above-mentioned problem we can use a naive method by using the Modulo (%) operator to find the remainder. But, the Modulo operator is computationally expensive and the method is inefficient.

Efficient Approach:
If we carefully observe the binary representation of N and its remainder with 4, we observe that remainder is simply the rightmost two bits in N. To get the rightmost two bits in number N, we perform bitwise AND (&) with 3 because 3 in binary is 0011. To understand the approach better let us have a look at the image below:

Below is the implementation of the above approach:

C

// C implementation to find N
// modulo 4 using Bitwise AND operator

#include <stdio.h>

// Function to find the remainder
int findRemainder(int n)
{

    // Bitwise AND with 3
    int x = n & 3;

    // return  x
    return x;
}

// Driver code
int main()
{

    int N = 43;
    int ans = findRemainder(N);

    printf("%d", ans);

    return 0;
}

C++

// C++ implementation to find N
// modulo 4 using Bitwise AND operator

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

// Function to find the remainder
int findRemainder(int n)
{
    // Bitwise AND with 3
    int x = n & 3;

    // Return  x
    return x;
}

// Driver code
int main()
{

    int N = 43;

    int ans = findRemainder(N);

    cout << ans << endl;

    return 0;
}

Java

// Java implementation to find N
// modulo 4 using Bitwise AND operator

class Main {

    // Driver code
    public static void main(String[] args)
    {

        int N = 43;

        int ans = findRemainder(N);

        System.out.println(ans);
    }

    // Function to find the remainder
    public static int findRemainder(int n)
    {
        // Bitwise AND with 3
        int x = n & 3;

        // return  x
        return x;
    }
}

Python 3

# Python 3 implementation to find N
# modulo 4 using Bitwise AND operator

# Function to find the remainder
def findRemainder(n):
    # Bitwise AND with 3
    x = n & 3

    # Return  x
    return x

# Driver code
if __name__ == '__main__':
    N = 43

    ans = findRemainder(N)

    print(ans)

# This code is contributed by Surendra_Gangwar

C#

// C# implementation to find N
// modulo 4 using Bitwise AND operator
using System;

class GFG {

    // Driver code
    public static void Main()
    {

        int N = 43;

        int ans = findRemainder(N);

        Console.Write(ans);
    }

    // Function to find the remainder
    public static int findRemainder(int n)
    {
        // Bitwise AND with 3
        int x = n & 3;

        // return  x
        return x;
    }
}

# This code is contributed by chitranayal

Javascript

<script>

// Javascript program implementation to find N
// modulo 4 using Bitwise AND operator

// Function to find the remainder
    function findRemainder(n)
    {
        // Bitwise AND with 3
        let x = n & 3;

        // return  x
        return x;
    }

// Driver Code

        let N = 43;

        let ans = findRemainder(N);

        document.write(ans);

</script>

Output

3

Time Complexity: O(1)
Auxiliary Space: O(1)

Another Approach:

We can solve this problem using  right shift operator (>>). The right shift operator shifts the bits of a number to the right by a specified number of positions. When we shift a number to the right by 2 positions (i.e., n >> 2), we effectively divide it by 4 and get the quotient as the result.

If we multiply the quotient by 4 and subtract it from the original number, we get the remainder. This can be expressed mathematically as:

N = 4 * (N >> 2) + (N & 3)

Below is the implementation of the above approach:

C++

// C++ implementation to find N
// modulo 4 using Bitwise AND operator

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

// Function to find the remainder
int findRemainder(int n)
{
    // find quotient using right shift operator
    int quotient = n >> 2;

    // find remainder using bitwise AND operator
    int remainder = n - (quotient << 2);

    return remainder;
}


// Driver code
int main()
{
    int N = 43;
    int ans = findRemainder(N);
    cout << ans << endl;

    return 0;
}

Java

public class Main {
    // Function to find the remainder
    public static int findRemainder(int n)
    {
        // find quotient using right shift operator
        int quotient = n >> 2;

        // find remainder using bitwise AND operator
        int remainder = n - (quotient << 2);

        return remainder;
    }

    // Driver code
    public static void main(String[] args)
    {
        int N = 43;
        int ans = findRemainder(N);
        System.out.println(ans);
    }
}
// This code is contributed by Prajwal Kandekar

Output

3

Time Complexity: O(1)

Auxiliary Space: O(1)