How about doing like below ?.
Assume we have the elements like {12, 3, 4, 15, 9, 9, 4, 12 }
Binary representation {1100, 0011, 0100, 1111, 1001, 1001, 0100, 1100 }
Step 1: Check the left most bit and separate them 0 and 1's.
{0100, 0011, 0100, 1111, 10001, 1001, 1100, 1100 }
Step 2: Now divide into two problems based on the number of 0's and 1's of
the bit that you have checked in the step1
{0100, 0011, 0100} {1111, 10001, 1001, 1100, 1100 }
Step 3: Take the next bit from the left and repeat the same like step1 and
continue
Thanks,
Sathaiah
On Tue, May 18, 2010 at 7:11 PM, CHERUVU JAANU REDDY <
jaanu.cher...@gmail.com> wrote:
>
> Here u r using extra space to store count values..
>
>
>
> ----------------------------------------
> CHERUVU JAANU REDDY
> M.Tech in CSIS
>
>
>
> On Tue, May 18, 2010 at 7:01 PM, Jagadish M <jagadis...@gmail.com> wrote:
>
>>
>> On May 18, 8:29 am, Terence <technic....@gmail.com> wrote:
>> > How do you maintain the heap? Could you explain in detail for the
>> > following example:
>> > 1 2 3 3 2 1 1 1 2 3 (n=10, k=3)
>>
>> Basically, in each node we maintain the key and its count.
>>
>> Initially, heap has the first element.
>> 1:1
>>
>> Search for 2 and insert( since its not found)
>> 1:1
>> /
>> 2:1
>>
>> Search for 3 and insert ( since its not found).
>>
>> 1:1
>> / \
>> 2:1 3:1
>>
>> Search for 3; since 3 is already there increment its count by 1.
>>
>> 1:1
>> / \
>> 2:1 3:2
>>
>> Search for 2; since 2 is already there increment its count by 1.
>>
>> 1:1
>> / \
>> 2:2 3:2
>>
>> and so on...
>>
>>
>> Since, there are only k distinct keys the heap size will at most be k;
>> so each search/insert/increment operation takes O(log k) time.
>>
>>
>> Jagadish
>> http://www.cse.iitb.ac.in/~jagadish<http://www.cse.iitb.ac.in/%7Ejagadish>
>>
>>
>>
>> >
>> > On 2010-5-17 22:38, Jagadish M wrote:
>> >
>> >
>> >
>> > > The best algorithm I can think of is to maintain a heap of k elements.
>> > > Takes O(n log k) time.
>> > > Is anything told about the values of the keys? If the keys have values
>> > > less than N, then it is possible to do
>> > > what you say, i.e sort them in place.
>> >
>> > > -Jagadish
>> > >http://www.cse.iitb.ac.in/~jagadish<http://www.cse.iitb.ac.in/%7Ejagadish>
>> >
>> > > On May 13, 7:06 pm, divya<sweetdivya....@gmail.com> wrote:
>> >
>> > >> This problem was asked in google phone interview. We have N element
>> > >> array with k distinct keys(No relation between k& N given so assume
>> > >> k<N)
>> > >> What is the best method to sort this array without using any extra
>> > >> memory? Please elaborate.
>> >
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