@Arun

Read Anup's comment "but It will go into O(n^2) for multiple occurrences of
multiple words."

On Thu, Aug 25, 2011 at 8:06 PM, Arun Vishwanathan
<aaron.nar...@gmail.com>wrote:

> @anup: can u provide a sort of pseudocode as to how ur code is O(n)?firstly
> u need to find out which word might have a repetition. so u compare first
> character of first word with all the other first characters.if there is not
> repetition , then u have to compare first character of second word with all
> the first character of words ahead of it and so on till u might sense a
> repetition.once u find a repetition u shift all the remaining words over the
> duplicated word.this is O(n) again.
> so are u doing this with a lot of pointers or how is that u keep it O(n)
> overall?
>
>
> On Thu, Aug 25, 2011 at 9:56 AM, sagar pareek <sagarpar...@gmail.com>wrote:
>
>> thanks ankur khurana
>>
>> @dave
>> May be u never did any mistake in posting and reading the problems.
>> But dont think urself superior.
>> Yeah i did mistake in reading the question so u must either ignore it or
>> request me to not repeat it in future.
>> You are behaving like its my daily routine of doing such kind of things.
>>
>>
>> On Thu, Aug 25, 2011 at 11:56 AM, Anup Ghatage <ghat...@gmail.com> wrote:
>>
>>> Actually, this method will be O(n) for any number of occurrences of a
>>> single word, but It will go into O(n^2) for multiple occurrences of multiple
>>> words.
>>>
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>>
>>
>>
>> --
>> **Regards
>> SAGAR PAREEK
>> COMPUTER SCIENCE AND ENGINEERING
>> NIT ALLAHABAD
>>
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>
>
>
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