@mukesh done :D
On Fri, Aug 12, 2011 at 1:07 PM, Mukesh kumar wrote:
> Hi all,
>
> i want to unsubscribe from this group.kindly make me out of this
> group. now i am feeling little iritation from these mails.so please
> ..do thid needful
>
>
> On 3/5/10, Umer Farooq wrote:
> > Thanks for let
@umar "u'll have to look back and see if the word already exists in the
list." if i am correct we need not to look up back , because when u will
insert words in dictionary (ordered or trie) , our algo will make sure that
if word already exist if yes increment corresponding count else insert node
Hi all,
i want to unsubscribe from this group.kindly make me out of this
group. now i am feeling little iritation from these mails.so please
..do thid needful
On 3/5/10, Umer Farooq wrote:
> Thanks for lettimg me know that. Although I have used this kinda thingy in
> solving a lot of proble
Thanks for lettimg me know that. Although I have used this kinda thingy in
solving a lot of problems; but, I never knew that it is called a trie.
On Thu, Mar 4, 2010 at 8:08 PM, Chakravarthi Muppalla
wrote:
> @Umer
> 'categorize the words into 26 categories depending on the initial
> character',
@Umer
'categorize the words into 26 categories depending on the initial character',
as far as i know this is the principle of a trie. look up trie for the next
word,
if exists increment count; other wise start counter and insert into trie; i
think this one would work.
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What about a ternary Search tree?
On 4 March 2010 16:21, Umer Farooq 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 expla
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;
.
trie data structure
On Sat, Feb 27, 2010 at 1:13 PM, subbu bvss wrote:
> i think u have to use t9 algorithm.. (tree type data structure)...
>
>
> On Sat, Feb 27, 2010 at 6:32 PM, abhijith reddy
> wrote:
>
>> You can use a TRIE .. Structure can be something like this
>>
>> struct trie
>> {
>>
i think u have to use t9 algorithm.. (tree type data structure)...
On Sat, Feb 27, 2010 at 6:32 PM, abhijith reddy 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
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
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 wrote:
You have to count the occurances of all words in a document. You are
given a method chat * GetNextWord, that returns the next wo
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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