@Dheeraj : I think u should review the problem again.
What u have posted is a way to find no. of set bits in a number.
But problem is check a bit at a position in all number for no. of 1s and 0s,
not in a single number.
This has to be done for all 32 bits.

Sanju
:)



On Sun, Aug 21, 2011 at 1:11 AM, Dheeraj Sharma <[email protected]
> wrote:

> while(a)
> (
> a&=(a-1)
> count++
> )
> counts number of 1s in number 'a'..
> Loop can be breaken if count exceeds 16..
>
> 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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