CareerCup

Good enough

int numbits(unsigned int n)
{
	for( int i=0; n > 0 ; i++ ) n &= (n-1);
	return i;
}

In Java:

public static int getNOnes(int n)
	{
		int result = 0;
		while(n>0)
		{
			n = n&(n-1);
			result++;
		}
	return result;
	}

#!/usr/bin/env python
def countOne(n): return n != 0 and 1 + countOne(n & (n-1)) or 0

This solution is quickrrrrrrr. even for all 16 bits it is O(8)

#include<stdio.h>
void main() {
   unsigned int num= 65535,no_of_bits=0;
   for( ;num>0 ; num = num>>2  )
      no_of_bits+=(num & 3)> 1 ? ((num & 3) - 1) : (num & 3 );
   printf("%d",no_of_bits);
}

/* Compile and Executed this code in Linux GCC 3.2*/

#include <stdio.h>

int count(int n)
{
int i=0;
if(n==0)
return 0;
else
{
while (n>0)
{
if(n&1)
++i;
n>>=1;
}
return i;
}
}

int main()
{
int n;
printf ("Enter no\n");
scanf ("%d",&n);
printf ("no of bit that are 1 one in this number is %d\n",count(n));
return 0;
}

public class shift {

public static void main(String str[]){
int a=20;
int b=1;
int count=0;;
for(int i=0;i<4*8;i++){
int c = b&a;
if(c>0){
count++;
}

b=b<<1;
}

System.out.println(count);
}
}

// Return bits that are 1 in the m 1-bit time
short return_bit_count(int num)
{
	short count = 0;
	while (num > 0)
	{
		++count;
		num &= (num-1);
	}
	return count;

}

int numOfOneBits(int number)
{
int count = 0, leftShift = 0;

while(number >= (1 << leftShift))
{
if(number & (1 << leftShift))
count++;

leftShift++;
}

return count;
}

void countOne(int n){
int cnt=0;
while(n>0){
if(n&1) cnt++;
n = n>>1;

}

void countOne(int n){
int cnt=0;
while(n>0){
n = n & (n-1);
cnt++;
}
printf("num is = %d", cnt);
}

int n=15;
	int count=0;
	int i=0;
	int log = 0;
	//compute the log2 of n(number of bynary chars)
	while (n >>= 1) ++log;
	n=15;
	for( i=0;i<=log;i++){
		if(n%2==1){
			count++;
		}
		n=n/2;
	}

	printf("%i",count);

this is for only 32 bit integer

#include<stdio.h>
int main()
{
int a=65535,i =0,tmp = 0;
while(i<32)
{
if(((a>>i)&1) == 1)
tmp++;
i++;
}
printf("total no of 1 is %d\n",tmp);

return 0;
}

for 32 bit ineteger
#include<stdio.h>
int main()
{
int a=65535,i =0,tmp = 0;
while(i<32)
{
if(((a>>i)&1) == 1)
tmp++;
i++;
}
printf("total no of 1 is %d\n",tmp);

return 0;
}

for 32 bit

#include<stdio.h>
int main()
{
int a=65535,i =0,tmp = 0;
while(i<32)
{
if(((a>>i)&1) == 1)
tmp++;
i++;
}
printf("total no of 1 is %d\n",tmp);

return 0;
}

int main(){
    unsigned int n = 34; // say this is the given number
    int count = 0;
    while(n){
        n &= (n-1);
        count++;
    }
    cout << count << endl;
    return 0;
}

One of the very basic implementations of the above algorithm is to use the & operator to & it with the number 1 to check if the current bit is 1 or not if yes then increment the count variable else skip the bit and then right shift the bit.

Implementation:

#include<bits/stdc++.h>
using namespace std;
void countbits(int x){
int count = 0;
while(x){
if(x & 1 == 1)
count++;
x = x >> 1;
}
return count;

}