@Vivek
I had told abt tat border case already once..
Suppose the two missin numbers are greater than n, then m==0 when exitin the
loop.
So they will be n+1 and n+2 only.
in case, one of the missin numbers is greater than n, then m==1, and can be
simply found by subtracting the (array_sum+x[0] )
The logic is actually simple. Tot if we mark in some way an element when
it's scanned, we can find the missing numbers in the second scannin.
3,5,1,2,9,10,8,6
When for loop sees '3' it knows elt 3 is there. So multiplies the number at
3rd position by some arbitrary number. (* I've taken the
for(i=1; i=n ;i++ ) //for ease of understanding starting the
array with 1.
{
if(a[ i ] n )
continue;
else
a[ a [ i ] ]*=n;
}
int x[2], m = 0;
for ( int i = 1; i =n ; i++)
{
if(a[ i ] % n == 0)
continue;
than n, then m==1, and can be
simply found by subtracting the (array_sum+x[0] ) from (sumof 1 to n+2)
numbers.
On Thu, Jul 30, 2009 at 10:55 PM, Devi G devs...@gmail.com wrote:
for(i=1; i=n ;i++ ) //for ease of understanding starting the
array with 1.
{
if(a[ i ] n
