@ankit...i m assuming that array of size n contains n-2 elements with 2
elements missing
On Mon, Jul 18, 2011 at 6:07 PM, SAMMM <somnath.nit...@gmail.com> wrote:
> #include<stdio.h>
> int main()
> {
>
> int a[]={1,2,3,5,2}; // 2 isrepeated and 4 is missing
> int i=0,x=0,j=0,bit;
> while(i<5)
> {
> j^=i+1;j^=a[i];
> i++;
> }
>
> bit=j&~(j-1); //set bits
>
> i=0;j=0;
>
> while(i<5)
> {
> if(bit&(i+1)) x^=(i+1); //two set are needed to iterate
> else j^=(i+1); //two set are needed to iterate
> i++;
> }
> i=0;
> while(i<5)
> {
> if(bit&a[i]) x^=a[i]; //two set are needed to iterate
> else j^=a[i]; //two set are needed to iterate
> i++;
> }
>
> printf("%d %d",x,j);
> return 0;
> }
>
> Hav a look .... The trick is in the set bit [ bit=j&~(j-1); //set
> bits]
>
> On Jul 18, 4:31 pm, TUSHAR_MCA <tusharkanta.r...@gmail.com> wrote:
> > Given an array of size n. It contains numbers in the range 1 to n.
> > Each number is present at least once except for 2 numbers. Find the
> > missing numbers ?
>
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