A little better way would be
x + y = p = (n*(n+1))/2 - (sum of all elements of array).
x * y  = q = n! / (product of all elements of the array).

now solve the quadratic eqn.
z^2 - (sum of the roots)z + (product of the roots)  = 0

On Mon, Jul 18, 2011 at 5:27 PM, varun pahwa <[email protected]>wrote:

> make two equations as .
> suppose numbers to be x,y
> x + y = p = (n*(n+1))/2 - (sum of all elements of array).
>
> x^2 + y^2 = q = (n*(n+1)*(2n+1))/6 - (sum of square of all elements of
> array).
>
> so 2*x*y can be calculated as (p^2 - q);
>
> so, a quad equation is formed as you now (x + y) and (2*xy).
>
> P.S. :: overflow is not handled.
>
> Please comment.
>
> On Mon, Jul 18, 2011 at 5:01 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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>
>
> --
> Varun Pahwa
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> 7th Sem.
> Indian Institute of Information Technology Allahabad.
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-- 
Regards,
Aashish Mann
Software Engineer
Vihaan Networks Ltd.,Gurgaon

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