Find number of ones in an integer.

Question

Jackson · Accepted Answer

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;
}

srikanth · Answer

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++;
}

Kitu · Answer

int numOfOnes( int  n ){

		int num = 0;

		String str = Integer.toString ( n );

		for(int i = 0 ; i < str.length() ;  i++ ) {

			if ( str.charAt ( i ) == '1' )
				num++;
                }

		return num;
	}

Harrison · Answer

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

Camus · Answer

#include 
#include 
using namespace std;

int CountBits(const int& number){
	int test = 1;
	int count = 0;
	for (int i = 0; i < (sizeof(number)*8) ; i++){
		int filter = number & test;

		if(filter)
			count++;	

		test <<= 1;
	}

	return count;
}

int _tmain(int argc, _TCHAR* argv[]){
	cout << CountBits(566632145) << endl;
	return 0;
}

pupkin · Answer

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

jinggle · Answer

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

Anonymous · Answer

Convert to binary and iterate. Check last big (increment counter if 1) then bit shift right to drop last bit. repeat.

Anonymous · Answer

int main(void)
{
    int input, counter = 0;
    printf("Enter integer: ");
    scanf("%d", &input);
    while(input)
    {
        if ((input & 0x0001) == 1)
            {counter++;}
        input = input >> 1;
    }
    printf("No. of 1s are: %d", counter);
}

Anonymous · Answer

Perl read fm stidn:

chomp ( my $int =  );
my $ones = $int =~ tr/1/1/;
say $ones;

DR · Answer

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.

jscott73 · Answer

In Perl:

sub OnesInNum{
    my $number = shift;
    my $count;

    my @numbers = split //, $number;
    foreach my $num (@numbers){
        next if($num eq '.'); #skip decimal point
        $count++ if($num == 1);
    }
    return $count;
}

DR · Answer

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;

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