@Vikram - sorry  to burst your bubble once again. try - [3 3 1 1]
even indirect xor'ing could not ensure anything useful.


On 2 November 2012 17:04, Vikram Pradhan <[email protected]> wrote:

> @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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