abt complexity.
now considering worst case scenario where swapping take place every time.
now assuming that we heapifying procedure , considering rows from i+1 to M
as M-(i+1) 1-dimensional array .
now heapify will take O(log(m*n)) time

so complexity would be M*N*log(M*N))

please correct me if i am doing wrong analysis.

and about the algorithm , i have little doubt if last row after completing
this procedure will remain sorted or not.

On Wed, Jan 11, 2012 at 9:42 PM, atul anand <[email protected]> wrote:

> @Lucifier :  it seem that the sub-matrix need to be heapifyed for A[i][j]
> is
>
>
> A[i+1][j] to A[row][col]
>
> there is no need to include A[i][j+1] to A[i][col] for A[i][j] as you have
> mentioned above.
>
> for eg :-
>
> 1  3   4   8     9              // 3 > 2 , so swap and heapify
>
> 2  5  18  25  50
> 6  7  22  45  55
>
> we get,
>
> 1  2   4   8     9
> 3  5  18  25  50
> 6  7  22  45  55
>
>
> replace it with 4 ( 4 > 3)
> we get,
>
> 1  2   3   8     9
> 4  5  18  25  50
> 6  7  22  45  55
>
> now after heapify this highlighted array we will get 4 as a min value ,
>  replace it with 8 ( 8 > 4)
>
> we get
>
> 1  2   3   4     9
> 8  5  18  25  50
> 6  7  22  45  55
>
> after heapifying we get,
>
> 1  2   3   4     9
> 5  8  18  25  50
> 6  7  22  45  55
>
> here 9 > 5 replace it.
>
> we, get
>
> 1  2   3   4     5
> 9  8  18  25  50
> 6  7  22  45  55
>
>
> after heapify we get,
>
>
> 1  2   3   4     5
> 6  8  18  25  50
> 9  7  22  45  55
>
> as we can see , 1st row is not included in the heapify procedure.
>
> correct me if i am wrong.
>
>
>
> On Wed, Jan 11, 2012 at 5:27 PM, Lucifer <[email protected]> wrote:
>
>> @Ankur..
>>
>> Yes... If you follow the algo that i have presented above and use
>> Atul's example you will be able to figure out..
>>
>> Maybe, the confusion is regarding heapfying.. ryt??
>> I am assuming so..
>>
>> Now as i said a submatrix rooted at A[i , j] is nothing but a heap
>> where its subtrees share a few nodes...
>>
>> Now, the first thing is to visualize the heap...
>> For a heap in the form of submatrix rooted at A[i][j], its children
>> are subtrees in the form of submatrix rooted at A[i+][j] and A[i][j
>> +1]...
>>
>> Now, imagine that you apply the normal heap-stabilizing approach when
>> the root element of a heap is being replaced by some other value...
>> Do it for an example submatrix (identified as explained above and also
>> whose rows and columns are sorted) and you will see how it works...
>>
>>
>> On Jan 11, 4:44 pm, Dipit Grover <[email protected]> wrote:
>> > @Shady : you would definitely need two index variables for each array I
>> > feel.
>>
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