@varun: can you write the code for one?

On Mon, Jul 18, 2011 at 5:11 AM, varun pahwa <[email protected]>wrote:

> The above solution will work if each other number is exactly once present.
> but if that 's not true.
> then 4 equations can be formed.
> Assuming a,b repeated number where a may or may be equal to b.
>
> then equations will be
> x + y = a + b;
> x^2 + y^2 = a^2 + b^2.
> x.y = a.b
> x^3 + y^3 = a^3 + b^3.
> now 4 equations 4 variables can be solved.
>
>
> On Mon, Jul 18, 2011 at 5:31 PM, ankit sambyal <[email protected]>wrote:
>
>> 1. Initialize a bit vector of size n.
>> 2. For every no. set the corresponding bit vector.
>> 3. Now scan through the bit vectors and get the missing numbers
>> corressponding to the unset bits in the bit vector.
>>
>>
>> Time complexity : O(n)
>>
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>
>
> --
> Varun Pahwa
> B.Tech (IT)
> 7th Sem.
> Indian Institute of Information Technology Allahabad.
> Ph : 09793899112
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>
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-- 
-Aakash Johari
(IIIT Allahabad)

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