let n be the no.of integers in the array :
int i=1,a;
int zero,one;
for(int a=1;a<=32;a++)
{
zero=0;
one=0;
for(int j=0;j<n;j++)
{
if(a[j] & i)
{
one++;
}
else
{
zero++;
}
}
if(one > zero)
{
printf("1s are more \n");
}
else
{
printf("0s are more \n");
}
i=i<<1;
}
Correct me if m wrong.
Sanju
:)
On Sun, Aug 21, 2011 at 1:28 AM, Dheeraj Sharma <[email protected]
> wrote:
> yeah i took it in the another way..i ll post it v soon
>
> On 8/21/11, himanshu kansal <[email protected]> wrote:
> > problem: There is an array containing integers.....
> > for every bit in the integer,you have to print a 1 if no of 1s
> > corresponding to that bit is more than no of 0s corresponding to that
> > bit (counting that bit in all the integers) otherwise print a 0(if no
> > of 0s corresponding to that bit are more).
> >
> > this you have to do for all bits in the integers.....
> >
> > assumption:integers are of 32bits.
> > no of integers in array are odd...(i.e. there is no case like no. of
> > 1s=no. of 0s)
> >
> > i have done this by counting the no of 1s and 0s for all bits.....
> >
> > but can anyone suggest any other efficient approach (somewhat using
> > bitwise operators).....
> >
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>
>
> --
> *Dheeraj Sharma*
> Comp Engg.
> NIT Kurukshetra
> +91 8950264227
>
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