Thanks dave.. :-)

On Thu, Mar 24, 2011 at 10:54 AM, Dave <[email protected]> wrote:
> @Ankit: You read in the first k numbers and form a max-heap of these
> numbers. This takes O(k) space. Then you read the rest of the file one
> number at a time. If the number is greater than the root of the max
> heap, then it is not one of the k smallest numbers. If it less than
> the root, replace the root with the smaller number and re-establish
> the heap condition. Thus, the memory requirement remains O(k). When
> you reach the end of the data, then the heap contains the k smallest
> numbers. If there are N numbers in the file, the work is O(N * log k).
>
> Dave
>
> On Mar 23, 11:22 pm, Ankit Sinha <[email protected]> wrote:
>> Guys,
>>
>> My intention was to understand how to manage max heap of k size into
>> memory. Means we have millions of numbers that we can't load into RAM
>> then how in the very first go we going to load only k size and how
>> will track of rest numbers. Can anybody code it? Do we need file to
>> store million numbers and then read say first k numbers and then take
>> another chunk?
>>
>> Thanks,
>>
>> Ankit!!
>>
>>
>>
>> On Tue, Mar 22, 2011 at 10:13 AM, Rajeev Kumar <[email protected]> 
>> wrote:
>> >http://flexaired.blogspot.com/2011/03/big-file-containing-billions-of...
>>
>> > On Mon, Mar 21, 2011 at 4:05 AM, Natansh Verma <[email protected]>
>> > wrote:
>>
>> >> @dave -was this a constraint since the beginning? In case it was, I am
>> >> sorry I didn't notice.
>>
>> >> In that case, the heap method ought to work better. I dont think the
>> >> quicksort method will work.
>>
>> >> Sent from my iPhone
>>
>> >> On 20-Mar-2011, at 23:00, Dave <[email protected]> wrote:
>>
>> >> > @Natansh: How do you do this with the constraint that your RAM is so
>> >> > small that you cannot accomodate all of the numbers at once?
>>
>> >> > Dave
>>
>> >> > On Mar 20, 9:04 am, Natansh Verma <[email protected]> wrote:
>> >> >> There's another way... use the partitioning method for quicksort to
>> >> >> find the
>> >> >> k smallest elements. Then it should take expected time as O(n + klogk).
>> >> >> Plus, it is in-place.
>>
>> >> >> On Wed, Mar 16, 2011 at 7:26 PM, asit <[email protected]> wrote:
>> >> >>> I agree with munna
>>
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>> > Rajeev Kumar
>>
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