@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 <[email protected]> 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 <[email protected]> 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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> SAGAR PAREEK
> COMPUTER SCIENCE AND ENGINEERING
> NIT ALLAHABAD
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