@Don :yeah sorry i misinterpreted the if condition ..it'll work fine .

I've modified my previous sol. and in this sol we do not need to modify the
array . Time complexity O(n) and constant space.
http://ideone.com/s2kR24





On Fri, Nov 2, 2012 at 1:14 AM, Don <[email protected]> wrote:

> It won't enter an infinite loop in that case. In fact, it would
> immediately return.
> Don
>
> On Oct 31, 2:39 pm, Vikram Pradhan <[email protected]> wrote:
> > @Don It will be an infinite loop for some cases  ...like try this i=1,
> and
> > a[1] = 5 , a[5] = 5
> >
> > *Solution:*
> >
> > As the numbers are from 0 to N-1 so we can xor the value with its index
> in
> > a loop . if the result is 0 then there is no repetition else there is
> some
> > repetition.
> >
> > *int result = 0;*
> > *for(int i=0;i<N;i++)*
> > *{*
> > *result ^= i ^ array[i];*
> > *}*
> > *
> > *
> > *if(result==0)*
> > *There is no repetition.*
> > *else*
> > *There is some repetition.*
> >
> > Time complexity O(N)
> > Space Complexity : constant
> >
> > check this :http://ideone.com/RXyynB
> >
> >  As the indexes are from 0 to N-1 and numbers are also from 0 to N-1 in
> > random order. So if there is no repetition then there is exactly two
> copies
> > of same number in set of (values and index) and when we xor all the
> indexes
> > to all the numbers the result will be zero because xor of same no. is
> zero.
> >
> > --
> > Vikram Pradhan | B.Tech| Computer Science & Engineering | NIT Jalandhar
>  |
> > 9740186063 |
>
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-- 
Vikram Pradhan | B.Tech| Computer Science & Engineering | NIT Jalandhar  |
9740186063 |

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