cant it be like '%f%'
On Mon, Sep 13, 2010 at 2:55 PM, Praveen Baskar <praveen200...@gmail.com>wrote:
> i think that we have to generate substrings from the given string such that
> the similarity is above 50%
> for eg.
> word ="foo"
> we have to generate the strings which must be greater than half of the
> given string length
> {"fo","oo"} (in this case)
>
> after this operation we have the following string set {"foo","fo","oo"}
>
> then we can do select* from product where name like '%foo%'....select*
> from product where name like '%fo%'..... select* from product where name
> like '%oo%'
>
>
> please do correct me if i am wrong
> On Mon, Sep 13, 2010 at 1:01 PM, SVIX
> <saivivekh.swaminat...@gmail.com>wrote:
>
>> select * from product
>> where
>> name like '%foo%'
>> and len(name) <=6;
>>
>> btw, how do u define similarity? i'm guessing it wouldn't be this
>> simple...
>>
>> On Sep 12, 5:12 am, Snoopy Me <thesnoop...@gmail.com> wrote:
>> > You are given the amazon.com database which consists of names of
>> > millions of products. When a user enters a search query for particular
>> > object with the keyword say "foo" , output all the products which have
>> > names having 50% or more similarity with the given keyword ie "foo"
>> >
>> > Write the most efficient algorithm for the same.
>>
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>
>
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> By B. Praveen
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