@Abhishek +1

On Wed, Jul 18, 2012 at 11:19 PM, Abhishek Sharma <abhi120...@gmail.com>wrote:

> sort each word in the list,then sort the whole list.
> Now,sort the input word(string).
> and then use binary search to find the word.
>
>
> On Wed, Jul 18, 2012 at 8:59 PM, Bhupendra Dubey 
> <bhupendra....@gmail.com>wrote:
>
>> 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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>>>
>>>
>>>
>>> --
>>> Vindhya Chhabra
>>>
>>>
>>>
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>>
>>
>>
>> --
>> Thanks & regards
>> Bhupendra
>>
>>
>>
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>
>
>
> --
> Abhishek Sharma
> Under-Graduate Student,
> PEC University of Technology
>
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