Assuming only one non-repeating element just run over the n elements and
keep xoring only the non repeated one will be left. O(1) space, O(n) time

On Sat, Apr 16, 2011 at 11:34 PM, sravanreddy001
<[email protected]>wrote:

> Assuming there is only one element which i not repeated,
>
> my approach will need O(n) space...
> load all distinct elements and they counts as you traverse them first..
> cost = O(n)
>
> searching an element from this.. O(n)
>
> any better memory management here(i mean space)
>
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B.Tech IV year CSE

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