yes this will work....

On 9/1/12, Navin Kumar <algorithm.i...@gmail.com> wrote:
> @atul: "Given a word and a dictionary find all the anagrams of that word in
> that dictionary". For efficient access i am storing each word in TRIE with
> their sorted key. Ex: "BAT" will be inserted with key ABT. By this we will
> form TRIE of all words.For all the words whose sorted version are  same ,
> they will reside at same node.For storing more than one word i am using
> linked list.
>
> Now for finding anagram of a given word i will just traverse with their
> corresponding sorted word and will display all the word at that node.
>
> Complexity: Each word insertion will take O(m),( sorting can be done in
> O(m)). If n words are in dictionary then total complexity will be:O(mn).
> This is the time complexity of building TRIE.
>
> Cost of displaying anagram of a word: O(m + d) where d = total output word
> .
>
> Comparison with HASH:
>
>  With this above approach we are saving time w.r.t. calculating hash key
> and  resolving collision time. Also writing a good hash function is also a
> challenge.
>
>
> On Sat, Sep 1, 2012 at 6:12 PM, atul anand <atul.87fri...@gmail.com> wrote:
>
>> yes but i would like to knw for which application you need this ?
>>
>> On 8/31/12, Carl Barton <odysseus.ulys...@gmail.com> wrote:
>> > There's no reason why a trie or a tree node couldn't be used to
>> 'represent'
>> > more than one word. Although you'd take a penalty in the complexity for
>> > searching etc.
>> >
>> > On 31 August 2012 15:33, Navin Kumar <algorithm.i...@gmail.com> wrote:
>> >
>> >> Can we store multiple words in single TRIE node by using linked list
>> >> or
>> >> some other data structure. Based on the some property a node in TRIE
>> will
>> >> hold all the word with same property.
>> >>
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