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

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)


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:
?
#include

/* Fnction 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 = 23;
if (isMultipleOf3(num))
printf("num is multiple of 3");
else
printf("num is not a multiple of 3");
getchar();
return 0;
}

Time Complexity O(logn)
Space Complexity O(1)
Source http://www.geeksforgeeks?211