This problem can be reduced if we are taking whole 32 bits...
Mean left most all 0's bits are also including
then if number is less than 65535 (2^16-1) then make it 0....
as 16 bits are at least zero in this case
On Sun, Aug 21, 2011 at 2:19 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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**Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
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