Hi Abhay,
Thanks for your solution, seems good as it is covering all the 3
attributes. Nesting of Hashmap to 3 layers is little clumsy. For Example if
we increase from 3 to 4 attributes we can see the complexity of handling
the things. I would like to know, do we have any other solutions.
On Mon, Apr 14, 2014 at 9:36 AM, Abhay Prakash <[email protected]>wrote:
> I think there is a correction for the time complexities:
>
> For the case when all three parameters are given, the time complexity will
> be *O(log(numTopics) + log(numLocations) + log(numLocale)). *Reason being
> it is like search on three different trees - after the key *topicValue*has
> been searched, only the tree associated to this key has to be searched
> for next parameter.
>
> Similar analysis for missing parameter(s) case will lead to:
> - when topicValue is missing - O(numTopics *
> (log(numLocation)+log(numLocale)) )
> - when locationValue is missing - O(log(numTopics) *+* numLocation
> ***log(numLocale)) *// note
> the + and * positions*
> - so on
>
> --
> Abhay Prakash
>
>
> On Sun, Apr 13, 2014 at 11:25 PM, Abhay Prakash
> <[email protected]>wrote:
>
>> Use a B-Tree implemented using nested map<> as:
>>
>> *map<string, map<string, map<string, yourResultDataType> > > table;*
>>
>> Now for each level the complexity will be log(numOfElementsAtThatLevel).
>> e.g.* if you have all three parameters, the complexity of retrieval will
>> be O(log(numTopics)*log(numLocations)*log(numLocale)).*
>>
>> For the missing parameters iterate on that level to get all possible
>> results with other two parameters. Iteration on that level will obviously
>> be O(numOfElementsOnThatLevel). e.g.* if only location is missing
>> complexity will be O(log(numTopics)*numLocations*log(numLocale))*
>>
>> ------ sample code --------
>> map<string, map<string, map<string, *yourResultDataType*> > > table;
>>
>> void addEntry(string topic, string location, string locale,
>> *yourResultDataType* rValue)
>> {
>> table*[topic][location][locale]* = rValue;
>> }
>>
>> yourResultDataType getResults(string topic, string location, string
>> locale)
>> {
>> return table[topic][location][locale]; // in case this entry doesn't
>> exist in table, *null *will be returned
>> }
>>
>> vector<yourResultDataType> getResults*WithoutTopic*(string location,
>> string locale) // return all possibilities with given two params.
>> {
>> vector<yourResultDataType> toReturn;
>> map<string, map<string, map<string, *yourResultDataType*> > >::
>> iterator it;
>> for(it = table.begin(); it != table.end(); ++it)
>> {
>> toReturn.push_back((it->second)*[location][locale]*);
>> }
>> return toReturn;
>> }
>>
>> vector<yourResultDataType> getResults*WithoutLocation*(string topic,
>> string locale)
>> {
>> vector<yourResultDataType> toReturn;
>> map<string, map<string, map<string, *yourResultDataType*> >
>> >::iterator ptr = table.find*(topic)*;
>> map<string, map<string, *yourResultDataType*> >:: iterator it;
>> for(it = ptr->second.begin(); it != ptr->second.end(); ++it)
>> {
>> toReturn.push_back(it->second*[locale]*);
>> }
>> return toReturn;
>> }
>>
>> ----- similarly for other combinations -------
>>
>> --
>> Abhay Prakash
>> IV Year (B.Tech + M.Tech Dual Degree)
>> Computer Science And Engineering
>> Indian Institute of Technology, Roorkee.
>>
>>
>> On Fri, Apr 11, 2014 at 10:34 AM, Narsimha Raju
>> <[email protected]>wrote:
>>
>>> Hi,
>>> I have a question and i would like to know the data structure to
>>> be used for this.
>>>
>>>
>>> You have 3 attribute
>>> 1. Topic
>>> 2. Location
>>> 3. Locale
>>>
>>>
>>> User can give any of the three or all the them to retrieve the content.
>>> For example.
>>>
>>> 1. Politics, Hyderabad, English
>>> In this case we need to show the result which satisfy all the
>>> 3 conditions.
>>>
>>> 2. Politics, Hyderabad
>>> In this case we need to show all the results of politics
>>> Hyderabad irrespective of language.
>>>
>>>
>>> --
>>> Regards,
>>> C. Narsimha Raju
>>>
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>>> an email to [email protected].
>>>
>>
>>
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--
Regards,
C. Narsimha Raju
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