Sort each of the word and form a trie , if any words comes again you get
one sch case.

On Wed, Jul 18, 2012 at 2:12 PM, vindhya chhabra
<vindhyachha...@gmail.com>wrote:

> yes,sorry count sort will be O(n) so better.thanks
>
> On Wed, Jul 18, 2012 at 1:43 PM, saurabh singh <saurab...@gmail.com>wrote:
>
>> ^sorting a string would be o(n^2logn) if u use q.sort.count sort would be
>> better.
>> Saurabh Singh
>> B.Tech (Computer Science)
>> MNNIT
>> blog:geekinessthecoolway.blogspot.com
>>
>>
>>
>> On Wed, Jul 18, 2012 at 1:08 PM, vindhya chhabra <
>> vindhyachha...@gmail.com> wrote:
>>
>>> sort the list,sort the word(use quick sort(nlogn  time))- and den search
>>> using binary search(logn time)
>>> or we can evn do by hashing-hash the word,den for every word keep
>>> decreasing the counter,if the hash array is zero ,anagram,else reset the
>>> hash array for given input for the checking the next word.
>>>
>>>
>>> On Wed, Jul 18, 2012 at 2:07 AM, Navin Kumar 
>>> <algorithm.i...@gmail.com>wrote:
>>>
>>>> Assuming a preexisting list of 100 words, how would you efficiently see
>>>> if a word received from input is an anagram of any of the 100 words?
>>>>
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>>>
>>>
>>> --
>>> Vindhya Chhabra
>>>
>>>
>>>
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>
>
>
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> Vindhya Chhabra
>
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-- 
Thanks & regards
Bhupendra

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