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Give a very good method to count the number of ones in a "n"
(e.g. 32) bit number.
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Answer / sujan
#include<iostream>
#define bit 32
using namespace std;
int array[bit];
int bitConvert(int n)
{
int a,j=0;
a=n%2;
for(int i=bit;i>=0;i--)
{
n=n/2;
array[i]=a;
a=n%2;
}
for(int i=0;i<=bit;i++)
{
cout<<array[i];
}
}
int countBit(int a[])
{
int *ptr;
ptr=a;
int j=0;
for(int i=0;i<=bit;i++)
{
if(*ptr==1)
{
j++;
}
ptr++;
}
cout<< j;
}
int main()
{
int n;
cout<<"Enter the no:";
cin>>n;
cout<<"\n"<<"BitConversion of "<<n<< "is:";
bitConvert(n);
cout<<endl<<endl;
cout<<"\n"<<"No. of bit:";
countBit(array);
system("pause");
}
| Is This Answer Correct ? | 4 Yes | 1 No |
Answer / artyom
// It's still O(n), maybe there are better ways.
int countBit(int num)
{
int count = 0;
while(num)
{
count += static_cast<int>(static_cast<bool>(mask &
0x1));
num >>= 1;
}
return count;
}
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / manish kumar
for (c = 0; n; c++)
n &= n - 1;
Result: the value of c.
complexity:o(logn)
| Is This Answer Correct ? | 1 Yes | 0 No |
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