@Harshal: I think ur code will print the input string in a sorted order.
@Snehi: Ur tree will never be balanced. and in worst case scenario there
will be only right child.so in that case generation of binary tree may go
upto O(n*n).
P.S.: correct me if i am wrong.

On Wed, Jun 22, 2011 at 1:50 PM, Rohit Sindhu <
[email protected]> wrote:

> @snehi .. your solution does not come upto the O(n) as for n elements of
> string it will take O(lg n) for each , so a total of O ( n  *  lg n )
>
> Otherwise a better variation to Solution is taking a count member in each
> node and incrementing it when another occurrence is made of that character.
>
>
> Correct me if i am wrong ,
>
>
> On Thu, Jun 23, 2011 at 12:51 AM, snehi jain <[email protected]>wrote:
>
>>  the binary tree for the above example will be
>>            k(1)
>>              \
>>                a(2)
>>               /  \
>>          (7) a     p(3)
>>                       \
>>                         i(4)
>>                          \
>>                            l(5)
>>                             \
>>                              r (6)
>>
>> number in the bracket denotes the order of insertion like k inserted first
>> then a then p
>> and so on ..
>> now if inorder is performed ... arent we getting  " kaapilr "
>>
>> On Thu, Jun 23, 2011 at 12:43 AM, oppilas . <[email protected]>wrote:
>>
>>> May be I didn't understood your logic.
>>> According to original question for
>>> I/P  ----kapilra
>>> O/P --kaapilr..
>>> Now,
>>> -what if we create a binary tree with root as the first element of the
>>> string and if the next character is equal then place it to left else place
>>> it to right. Similar comparison will be done while inserting all the other
>>> nodes too .
>>>
>>>    At root you will insert k(count=1)
>>> After that a
>>>
>>>                      k(1)
>>>                    /
>>>                 a(1)
>>> Then you will insert p,
>>>                      k(1)
>>>                    /      \
>>>                  a(1)    p(1)
>>>  And so on.
>>>
>>>>
>>>>> after that if InOrder traversal is performed.. it would give us the
>>>>> desired output.
>>>>>
>>>>
>>> If by inoder traversal we can get desired output then please how me how
>>> using a small example :).
>>>
>>>
>>>>
>>>>> Snehi
>>>>>
>>>>>
>>>>> On Wed, Jun 22, 2011 at 9:48 PM, DK <[email protected]> wrote:
>>>>>
>>>>>> No. This is equivalent to a sort with comparisons based on index of
>>>>>> first occurrence in the input string. Any comparative algorithm is O(n 
>>>>>> log
>>>>>> n) and a non comparative algorithm can be O(n) only by using counting or
>>>>>> radix sorting etc.
>>>>>>
>>>>>> --
>>>>>> DK
>>>>>>
>>>>>> http://twitter.com/divyekapoor
>>>>>> http://www.divye.in
>>>>>>
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>
>
>
>
> Regards ,
>
> --
> Rohit Sindhu
>
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-- 
Varun Pahwa
B.Tech (IT)
7th Sem.
Indian Institute of Information Technology Allahabad.
Ph : 09793899112 ,08011820777
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