Interview Question for Software Engineer at Apple:

1. Convert the number into bits (we assume that all our bits of size 8)
2. "And" it with 1
3. check if the bit returned is 1 or now. if it is '1' increase some counter.
4. right shift 8 times and perform go to 2

or you can do the following using properties of number.
every number when AND'ed with number - 1 it would eliminate one binary number.\

count = 0;
while(n) {
   n &= (n-1);
   count++;
}

This doesn't work:

1. Convert the number into bits (we assume that all our bits of size 8)
2. "And" it with 1
3. check if the bit returned is 1 or now. if it is '1' increase some counter.
4. right shift 8 times and perform go to 2

look at the number 11:

counter=0;
0x000B, look at the last byte B: 1011
and that with 1, you get 1
counter now equals 1, right shift 8 times and get 0x0000
and with 1, get 0, doesn't increment counter.

so the number of 1's in 11 is 1.

This doesn't work either:

or you can do the following using properties of number.
every number when AND'ed with number - 1 it would eliminate one binary number.\

count = 0;
while(n) {
   n &= (n-1);
   count++;
}

Look at n=7, which we know has no 1's in it.

7 in binary: 0111
6 in binary: 0110

n=7&6;
so n=6;
counter=1;

loop again:
5 in binary=0101;

n=6&5=0110&0101=0100=4;
counter=2;

We can already see this is wrong answer should be 0;

Solutions provided above are used for finding out the number of ones in a binary string. If you Google for such question, both proposed answers would show up.

For finding out the number of ones in an integer, I would propose the following.

int NumberOfOnes(double number)
{
    int counter = 0;
    if (number == 0) return 0;
    if (number == 1) return 1;

    do
    {
        if (number % 10 == 1) counter++;
        number = number / 10;
    }
    while (number);

    return counter;
}

Binary ones in python:

def findBinOnes(integer):
    val = integer
    ones = 0
    while val > 0:
        if val & 1 == 1:
            ones +=1
        val = val >> 1
    return ones

Decimal ones in python:
def findDecOnes(integer):
    val = integer
    ones = 0
    while val > 0:
        if val % 10 == 1:
            ones += 1
        val = val / 10
    return ones

just an idea

#!/usr/bin/env ruby
123456718791.to_s.scan(/1/).size