I'm sorry...... misunderstood the problem and tot the missin numbers are to
be less than n.
Yet it can be modified to suit the pbm.
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] ) 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 )
> continue;
> else
> a[ a [ i ] ]*=n;
> }
>
> int x[2], m = 0;
>
> for ( int i = 1; i <=n ; i++)
> {
> if(a[ i ] % n == 0)
> continue;
> else
> {
> x[m++ ] = i;
> if(m == 2) break;
> }
> }
>
> x[] now contain the two missing numbers in ascending order.
>
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