Write an Efficient Method to Check if a Number is Multiple of 3

The very first solution that comes to our mind is the one that we learned in school. If sum of digits in a number is multiple of 3 then number is multiple of 3 e.g., for 612 sum of digits is 9 so it’s a multiple of 3. But this solution is not efficient. You have to get all decimal digits one by one, add them and then check if sum is multiple of 3.



There is a pattern in binary representation of the number that can be used to find if number is a multiple of 3. If difference between count of odd set bits (Bits set at odd positions) and even set bits is multiple of 3 then is the number.

Example : 23 (00..10111)
1) Get count of all set bits at odd positions (For 23 it’s 3).
2) Get count of all set bits at even positions (For 23 it’s 1).
3) If difference of above two counts is a multiple of 3 then number is also a multiple of 3.

(For 23 it’s 2 so 23 is not a multiple of 3)

Take some more examples like 21, 15, etc…

Algorithm: isMutlipleOf3(n)
1) Make n positive if n is negative.
2) If number is 0 then return 1
3) If number is 1 then return 0
4) Initialize: odd_count = 0, even_count = 0
5) Loop while n != 0
    a) If rightmost bit is set then increment odd count.
    b) Right-shift n by 1 bit
    c) If rightmost bit is set then increment even count.
    d) Right-shift n by 1 bit
6) return isMutlipleOf3(odd_count - even_count)

Proof:
Above can be proved by taking the example of 11 in decimal numbers. (In this context 11 in decimal numbers is same as 3 in binary numbers)
If difference between sum of odd digits and even digits is multiple of 11 then decimal number is multiple of 11. Let’s see how.

Let’s take the example of 2 digit numbers in decimal
AB = 11A -A + B = 11A + (B – A)
So if (B – A) is a multiple of 11 then is AB.

Let us take 3 digit numbers.

ABC = 99A + A + 11B – B + C = (99A + 11B) + (A + C – B)
So if (A + C – B) is a multiple of 11 then is (ABC)

Let us take 4 digit numbers now.
ABCD = 1001A + D + 11C – C + 999B + B – A
= (1001A – 999B + 11C) + (D + B – A -C )
So, if (B + D – A – C) is a multiple of 11 then is ABCD.

This can be continued for all decimal numbers.
Above concept can be proved for 3 in binary numbers in the same way.

Time Complexity:
O(logn)

Program:

C++

// CPP program to check if n is a multiple of 3
#include<bits/stdc++.h>
using namespace std;

/* Function to check if n is a multiple of 3*/
int isMultipleOf3(int n)
{
    int odd_count = 0;
    int even_count = 0;

    /* Make no positive if +n is multiple of 3
    then is -n. We are doing this to avoid
    stack overflow in recursion*/
    if(n < 0) n = -n;
    if(n == 0) return 1;
    if(n == 1) return 0;

    while(n)
    {
        /* If odd bit is set then
        increment odd counter */
        if(n & 1) 
        odd_count++;
        n = n>>1;

        /* If even bit is set then
        increment even counter */
        if(n & 1)
            even_count++;
        n = n>>1;
    }

    return isMultipleOf3(abs(odd_count - even_count));
}

/* Program to test function isMultipleOf3 */
int main()
{
    int num = 24;
    if (isMultipleOf3(num)) 
        printf("%d is multiple of 3", num);
    else
        printf("%d is not a multiple of 3", num);
    return 0;
}

Java

// Java program to check if
// n is a multiple of 3
import java.lang.*;
import java.util.*;

class GFG
{
 // Function to check if n 
 // is a multiple of 3
 static int isMultipleOf3(int n)
  {
    int odd_count = 0;
    int even_count = 0;

    /* Make no positive if +n is multiple 
    of 3 then is -n. We are doing this to 
    avoid stack overflow in recursion*/
    if(n < 0) n = -n;
    if(n == 0) return 1;
    if(n == 1) return 0;

    while(n != 0)
    {
        /* If odd bit is set then
        increment odd counter */
        if((n & 1) != 0 ) 
        odd_count++;
        n = n>>1;

        /* If even bit is set then
        increment even counter */
        if((n & 1) != 0)
            even_count++;

        n = n>>1;
    }

    return isMultipleOf3
    (Math.abs(odd_count - even_count));
}
    
/* Program to test function isMultipleOf3 */
    public static void main(String[] args) 
    {
    int num = 24;

    if (isMultipleOf3(num) != 0) 
        System.out.println(num + 
        " is multiple of 3");
    else
        System.out.println(num + 
        " is not a multiple of 3");
}
}

// This code is contributed by Sahil_Bansall

Python3

# Python profram to check if n is a multiple of 3

# Function to check if n is a multiple of 3
def isMultipleOf3(n):

    odd_count = 0
    even_count = 0

    # Make no positive if +n is multiple of 3
    # then is -n. We are doing this to avoid
    # stack overflow in recursion
    if(n < 0): 
        n = -n
    if(n == 0):
        return 1
    if(n == 1): 
        return 0

    while(n):
        
        # If odd bit is set then
        # increment odd counter 
        if(n & 1): 
            odd_count += 1
        n = n >> 1

        # If even bit is set then
        # increment even counter 
        if(n & 1):
            even_count += 1
        n = n >> 1

    return isMultipleOf3(abs(odd_count - even_count))

# Program to test function isMultipleOf3 
num = 24
if (isMultipleOf3(num)): 
    print(num, 'is multiple of 3')
else:
    print(num, 'is not a multiple of 3')

# This code is contributed by Danish Raza

C#

// C# program to check if
// n is a multiple of 3
using System;

class GFG {
    
    // Function to check if n
    // is a multiple of 3
    static int isMultipleOf3(int n)
    {
        int odd_count = 0, even_count = 0;

        /* Make no positive if +n is multiple 
        of 3 then is -n. We are doing this to 
        avoid stack overflow in recursion*/
        if (n < 0) n = -n;
        if (n == 0) return 1;
        if (n == 1) return 0;

        while (n != 0) {
            
            /* If odd bit is set then
            increment odd counter */
            if ((n & 1) != 0)
                odd_count++;
                
            n = n >> 1;

            /* If even bit is set then
            increment even counter */
            if ((n & 1) != 0)
                even_count++;

            n = n >> 1;
        }

        return isMultipleOf3(Math.Abs(odd_count - even_count));
    }

    // Driver Code
    public static void Main()
    {
        int num = 24;

        if (isMultipleOf3(num) != 0)
            Console.Write(num + " is multiple of 3");
        else
            Console.Write(num + " is not a multiple of 3");
    }
}

// This code is contributed by Sam007

PHP

<?php
// PHP program to check if n
// is a multiple of 3


// Function to check if n 
// is a multiple of 3
function isMultipleOf3( $n)
{
    $odd_count = 0;
    $even_count = 0;

    // Make no positive if +n 
    // is multiple of 3 then is -n. 
    // We are doing this to avoid
    // stack overflow in recursion
    if($n < 0) $n = -$n;
    if($n == 0) return 1;
    if($n == 1) return 0;

    while($n)
    {
        // If odd bit is set then
        // increment odd counter 
        if($n & 1) 
        $odd_count++;
        $n = $n>>1;

        // If even bit is set then
        // increment even counter
        if($n & 1)
            $even_count++;
        $n = $n >> 1;
    }

    return isMultipleOf3(abs($odd_count - 
                             $even_count));
}

// Driver Code
$num = 24;
if (isMultipleOf3($num)) 
    echo $num, "is multiple of 3";
else
    echo $num,"is not a multiple of 3";
    
// This code is contributed by vt_m.
?>


Output :

24 is multiple of 3

Asked in: amazon, microsoft

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