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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