That solution is using the sign bit as extra storage, which is clever,
but if you have an array of unsigned integers and N is more than half
of the largest unsigned integer, it won't work. There is a way to do
it without using the sign bit as extra storage.

On Oct 30, 5:16 pm, rahul sharma <[email protected]> wrote:
> i thnik this is the solution...http://www.geeksforgeeks.org/archives/9755
>
>
>
>
>
>
>
> On Wed, Oct 31, 2012 at 2:20 AM, Don <[email protected]> wrote:
> > We can modify the array. The algorithm should work even if we use
> > unsigned integers and N is the largest unsigned integer.
>
> > Don
>
> > On Oct 30, 4:42 pm, rahul sharma <[email protected]> wrote:
> > > Can we modify the array???we can make index we visit as negative and then
> > > if any one already containing -ve..then its repeating
>
> > > On Wed, Oct 31, 2012 at 1:40 AM, Don <[email protected]> wrote:
> > > > Does your algorithm work if N=4 and the array is {1,1,2,2}.
>
> > > > Don
>
> > > > On Oct 30, 2:32 pm, arumuga abinesh <[email protected]> wrote:
> > > > > if sum of all elements = n(n-1)/2 , no elements are repeated
> > > > > else some numbers are repeated
>
> > > > > On Tue, Oct 30, 2012 at 11:57 PM, Don <[email protected]> wrote:
> > > > > > Given an array of N integers in the range 0..N-1, determine if any
> > > > > > number is repeated in the array.
> > > > > > Solution should execute in O(n) time and use constant space.
> > > > > > Don
>
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