Well I like Samm's approach.Another possible approach would be adding all
the numbers.The result should fit in long long(2^32
numbers*max_int=2^64).Precompute the sum of 2^64 numbers by big int
arithematic.Subtract to get the missing numbers.Again o(1) space and o(n)
time though slower than SAMM's approach.

PS:Though I still doubt the problem requires the same.Dumanshu can u quote
the source of the problem please?

On Mon, Jul 18, 2011 at 12:06 PM, ankit sambyal <[email protected]>wrote:

> @lalit : I wanted to use bit vector to find the missing no. since the
> list contains all the numbers from 0 to 2^32 except one missing no.
> But it would require a bit vector of 2^32 bits which won't fit in
> memory. So, I found the narrowed the range in which the missing no.
> lies and then applied the bit vector concept. I hope its clear now ...
>
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-- 
Saurabh Singh
B.Tech (Computer Science)
MNNIT ALLAHABAD

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