Efficiently check whether n is a multiple of 4 or not

Given a number n. The problem is to efficiently check whether n is a multiple of 4 or not without using arithmetic operators.

Examples:

Input : 16
Output : Yes

Input : 14
Output : No

Approach: A multiple of 4 always has 00 as its last two digits in its binary representation. We have to check whether the last two digits of n are unset or not.

How to check whether the last two bits are unset or not.
If n & 3 == 0, then the last two bits are unset, else either both or one of them are set.

C/C++

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// C++ implementation to efficiently check whether n
// is a multiple of 4 or not
#include <bits/stdc++.h>
using namespace std;
  
// function to check whether 'n' is
// a multiple of 4 or not
string isAMultipleOf4(int n)
{
    // if true, then 'n' is a multiple of 4
    if ((n & 3) == 0)
        return "Yes";
  
    // else 'n' is not a multiple of 4
    return "No";
}
  
// Driver program to test above
int main()
{
    int n = 16;
    cout << isAMultipleOf4(n);
    return 0;
}

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Java

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// Java implementation to efficiently check
// whether n is a multiple of 4 or not
  
class GFG {
    // method to check whether 'n' is
    // a multiple of 4 or not
    static boolean isAMultipleOf4(int n)
    {
        // if true, then 'n' is a multiple of 4
        if ((n & 3) == 0)
            return true;
  
        // else 'n' is not a multiple of 4
        return false;
    }
  
    // Driver method
    public static void main(String[] args)
    {
        int n = 16;
        System.out.println(isAMultipleOf4(n) ? "Yes" : "No");
    }
}

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Python 3

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# Python 3 implementation to 
# efficiently check whether n
# is a multiple of 4 or not
  
# function to check whether 'n'
# is a multiple of 4 or not
def isAMultipleOf4(n):
  
    # if true, then 'n' is
    # a multiple of 4
    if ((n & 3) == 0):
        return "Yes"
  
    # else 'n' is not a
    # multiple of 4
    return "No"
  
# Driver Code
if __name__ == "__main__":
  
    n = 16
    print (isAMultipleOf4(n))
  
# This code is contributed 
# by ChitraNayal

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C#

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// C# implementation to efficiently check
// whether n is a multiple of 4 or not
using System;
  
class GFG {
      
    // method to check whether 'n' is
    // a multiple of 4 or not
    static bool isAMultipleOf4(int n)
    {
        // if true, then 'n' is a multiple of 4
        if ((n & 3) == 0)
            return true;
  
        // else 'n' is not a multiple of 4
        return false;
    }
  
    // Driver method
    public static void Main()
    {
        int n = 16;
        Console.WriteLine(isAMultipleOf4(n) ? "Yes" : "No");
    }
}
  
// This code is contributed by vt_m.

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PHP

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<?php
// PHP implementation to 
// efficiently check whether n
// is a multiple of 4 or not
  
// function to check whether 'n' is
// a multiple of 4 or not
function isAMultipleOf4($n)
{
      
    // if true, then 'n' 
    // is a multiple of 4
    if (($n & 3) == 0)
        return "Yes";
  
    // else 'n' is not 
    // a multiple of 4
    return "No";
}
  
    // Driver Code
    $n = 16;
    echo isAMultipleOf4($n);
      
// This code is contributed by anuj_67. 
?>

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Output:

Yes

Can we generalize above solution?
Similarly we can check for other powers of 2. For example, a number n would be multiple of 8 if n & 7 is 0. In general we can say.

// x must be a power of 2 for below
// logic to work
if (n & (x -1) == n)
   n is a multiple of x
Else
   n is NOT a multiple of x

This article is contributed by Ayush Jauhri. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : vt_m, Ita_c



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