Considering sequence 1, 2, 3, 4, 4 and n = 5
lets have missing number as X and repeated number as Y.

(1*2*3*4*4)/n! = 4/5=Y/X => 5Y = 4X => Y=4X/5

(sum of all numbers) + X - Y = n(n+1)/2.
14 + X - Y = 15
X-4X/5=1

X = 5 ---> missing value is 5.
Y = 4 --> repeated value.



On Wed, Oct 20, 2010 at 7:26 PM, Dave <dave_and_da...@juno.com> wrote:

> @Mahesh: Let's try this on 1, 2, 3, 4, 4. Then S = 14 and X_SUM = 0.
> But the duplicate element is 4, not (14-0) / 2 = 7.
>
> Dave
>
> On Oct 20, 5:49 am, Mahesh_JNU <mahesh.jnumc...@gmail.com> wrote:
> > Just add the number of the array and let the sum is S. Its complexity is
> > O(n).
> > Now XOR all elements of the array and say the result is X_SUM.Its
> complexity
> > is  O(n).
> > Now the duplicate element is = (S - X_SUM)/2
> >
> > On Wed, Oct 20, 2010 at 4:14 PM, Asquare <anshika.sp...@gmail.com>
> wrote:
> > > @Anirvana - In context to the XOR method u suggested, could u plz
> > > explain why does it so happen.. ??
> >
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-- 
Thanks & Regards,
Karthik

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