#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 <[email protected]> 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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