question is in c++

On Fri, Jun 3, 2011 at 7:48 AM, sanjay ahuja <[email protected]>wrote:

> you can use Java HashTable/HashMap to store key value pair, where Key
> being the element stored in array and Value being the index of the
> element. Any duplicate insert on key will replace the value of that
> Key. After finishing all elements retrieve all Key, value pair stored
> in HashMap.
>
> Let me know If I am missing something!!
>
> But if you want to make your own hashtable. Then we can find the range
> of elements in one pass and then carefully choose some hash function
> to avoid collision and make hash on key. But this will not be one
> pass.
>
> On Fri, Jun 3, 2011 at 7:31 AM, D.N.Vishwakarma@IITR <[email protected]>
> wrote:
> > @Sanjay ahuja
> > yes ans is required in O(n)
> > how it can be done by hashing?
> >
> > On Fri, Jun 3, 2011 at 1:19 AM, Supraja Jayakumar <
> [email protected]>
> > wrote:
> >>
> >> Hi
> >>
> >> I think multi hash table will solve this problem.
> >>
> >> Thanks
> >> Supraja J
> >>
> >> On Thu, Jun 2, 2011 at 1:37 PM, Harshal <[email protected]> wrote:
> >>>
> >>> But I think the solution required is O(n) (one pass). So, O(nlogn) is
> not
> >>> what the author wants anyway. Hashing is an option but we dont know the
> >>> range of the elements in the array. So, in case when all keys hash into
> the
> >>> same place, the worst case is still O(n^2).
> >>> suppose our hash function is h(n) = n mod 10.
> >>> array: 2,12,32,52,42.
> >>>
> >>>
> >>> On Fri, Jun 3, 2011 at 1:01 AM, Harshal <[email protected]> wrote:
> >>>>
> >>>> you can check if the element to be inserted next is same as the node
> >>>> value while inserting itself. Why to do a separate inorder traversal..
> >>>>
> >>>> On Fri, Jun 3, 2011 at 12:56 AM, Supraja Jayakumar
> >>>> <[email protected]> wrote:
> >>>>>
> >>>>> Hi
> >>>>>
> >>>>> Inserting into BST and break on same element will detect duplicates.
> >>>>> But in any case, this is still O(2n) ? - once for inserting them in
> to
> >>>>> the tree
> >>>>> and another for the inorder read.
> >>>>>
> >>>>> Correct me if wrong.
> >>>>>
> >>>>> Rgds
> >>>>> Supraja J
> >>>>>
> >>>>> On Thu, Jun 2, 2011 at 1:11 PM, Harshal <[email protected]> wrote:
> >>>>>>
> >>>>>> @Anurag: XOR wont work here, 1 element is repeated, not 1 element is
> >>>>>> unique. Read the question again.
> >>>>>> Keep inserting elements in a BST and break once you find the same
> >>>>>> element. O(nlogn)
> >>>>>>
> >>>>>> On Fri, Jun 3, 2011 at 12:38 AM, Anurag Narain
> >>>>>> <[email protected]> wrote:
> >>>>>>>
> >>>>>>> take X-OR of all the elements.....the one which has no duplicate
> will
> >>>>>>> be left and rest all will be reduced to zero.
> >>>>>>>
> >>>>>>> --
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> >>>>>>
> >>>>>>
> >>>>>>
> >>>>>> --
> >>>>>> Harshal Choudhary,
> >>>>>> III Year B.Tech CSE,
> >>>>>> NIT Surathkal, Karnataka, India.
> >>>>>>
> >>>>>>
> >>>>>> --
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> >>>>>
> >>>>>
> >>>>>
> >>>>> --
> >>>>> U
> >>>>>
> >>>>> --
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> >>>>
> >>>>
> >>>>
> >>>> --
> >>>> Harshal Choudhary,
> >>>> III Year B.Tech CSE,
> >>>> NIT Surathkal, Karnataka, India.
> >>>>
> >>>>
> >>>
> >>>
> >>>
> >>> --
> >>> Harshal Choudhary,
> >>> III Year B.Tech CSE,
> >>> NIT Surathkal, Karnataka, India.
> >>>
> >>>
> >>> --
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> >>
> >>
> >>
> >> --
> >> U
> >>
> >> --
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> >
> >
> >
> > --
> > With Regards
> > Deoki Nandan Vishwakarma
> > IITR MCA
> > Mathematics Department
> >
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> >
>
>
>
> --
> Sanjay Ahuja,
> Analyst, Financing Prime Brokerage
> Nomura Securities India Pvt. Ltd
>
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>


-- 
**With Regards
Deoki Nandan Vishwakarma
IITR MCA
Mathematics Department*
*

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