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++ 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; 
} 

// 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");
    }
}

# 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

// 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.

<?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. 
?>

<script>

// Javascript implementation to efficiently check
// whether n is a multiple of 4 or not
    
    // method 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 true;
  
        // else 'n' is not a multiple of 4
        return false;
    }
    
    // Driver method
    let n = 16;
    document.write(isAMultipleOf4(n) ? "Yes" : "No");
    
    //This code is contributed by rag2127
    
</script>

Output: 
 

Yes

Time Complexity : O(1)

Auxiliary Space: O(1)

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) == 0)
   n is a multiple of x
Else
   n is NOT a multiple of x