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

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// 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;
}

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C++














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// 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; 


} 



















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Java














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// 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; 


    } 


} 



















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Python 3














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# 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 



















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C#














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// 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 



















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Output:


3





Time Complexity: O(1)


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