It would work i guess
On Aug 21, 1:49 pm, Sanjay Rajpal <srn...@gmail.com> wrote:
> 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 <dheerajsharma1...@gmail.com
>
> > wrote:
> > yeah i took it in the another way..i ll post it v soon
>
> > On 8/21/11, himanshu kansal <himanshukansal...@gmail.com> 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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