What about a ternary Search tree?
On 4 March 2010 16:21, Umer Farooq <the.um...@gmail.com> wrote:
> I can't get how will u manipulate the trie DS. u'll have to look back and
> see if the word already exists in the list. this will be an extra overhead.
>
> I have thought of another algorithm. Here is an abstract explanation:
>
> The node used will be like this
> struct Word {
> .........char *_word;
> .........int _freq;
> .........Word *Next;
> }
>
> class Category {
> ........Word *List;
> ..... bool addWord(char *word); // if the word exists in the list, then
> increment the freqency, else add the word
> ........bool doesExist(word);
> }
>
> public void main (String[] args)
> {
> .... Category _cat[26];
> .....while (!EOF()) {
> ............char *c = getNextWord();
> ............_cat[c[0].toInt()-97].addWord(c);
> }
>
> Categorize the words into 26 categories depending on the initial character;
>
> Depending on the initial character, traverse the list of corresponding
> category
> If the word exists, then increment the counter of that word
> else add the word in the list of that category.
> On Thu, Mar 4, 2010 at 6:51 AM, ankur aggarwal
> <ankur.mast....@gmail.com>wrote:
>
>> trie data structure
>>
>>
>> On Sat, Feb 27, 2010 at 1:13 PM, subbu bvss <bvss.su...@gmail.com> wrote:
>>
>>> i think u have to use t9 algorithm.. (tree type data structure)...
>>>
>>>
>>> On Sat, Feb 27, 2010 at 6:32 PM, abhijith reddy <
>>> abhijith200...@gmail.com> wrote:
>>>
>>>> You can use a TRIE .. Structure can be something like this
>>>>
>>>> struct trie
>>>> {
>>>> int count; // no of occurences
>>>> char *child[SIZE];
>>>> };
>>>>
>>>> when ever u insert ( it will take just O(length) time) .. just increment
>>>> count by 1
>>>>
>>>> For each query (also O(length) time) the no of occurrences of the word
>>>> will be count of the last character
>>>>
>>>> Hope it helps
>>>>
>>>>
>>>>
>>>> On Sat, Feb 27, 2010 at 5:35 PM, <piyushgoe...@gmail.com> wrote:
>>>>
>>>>> Maintain a hash of word to freq. Keep adding words and incrementing
>>>>> their frequencies while reading the documents
>>>>>
>>>>>
>>>>> Pigol
>>>>>
>>>>>
>>>>> On Feb 27, 2010, at 5:10 PM, vijay <auvija...@gmail.com> wrote:
>>>>>
>>>>> You have to count the occurances of all words in a document. You are
>>>>>> given a method chat * GetNextWord, that returns the next word from the
>>>>>> document.
>>>>>> - Which datastructure can be userd to achieve this
>>>>>> -
>>>>>>
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>>
>>
>>
>> --
>> With Regards
>>
>> Ankur Aggarwal
>> Research internee
>> Optical Zeitgeist Laboratory
>> Institut National de la Recherche Scientifique (INRS) - ÉMT
>> 800, de la Gauchetière Ouest, bureau 6900
>> Montréal, QC, H5A 1K6
>> CANADA
>>
>> Ph: +1 514 966-2661
>> E-mail: ankur.1...@gmail.com
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>>
>>
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