@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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