@anurag
I meant the sorting without using any auxiliary data structure .Also  we
have to put the element in the tree and carry out a traversal for every
element we insert in the tree .This takes O(n*logn) time ,
If one can sort the elements using a stack in O(n) time , we better sort
with this method , say "Stack sort"
Moral:-No (comparison based )sorting method exists for which time complexity
is less than O(n*log n)
also , without using any auxialliary data structure , we cannot create all
the permutations of the stack
Refer Donald knuth's TAOCP for more details
On Mon, Jun 14, 2010 at 9:18 AM, Anurag Sharma <anuragvic...@gmail.com>wrote:

> Why not just pop all elements from stack ( O(n) )  and insert it in a self
> balancing Binary Search Tree (like RB Tree) (O(log(n) ) and then do and
> inorder traversal ( O(n) )and push elements in stack again.
> Time = O(nlog(n) + n)
> Space=O(n) (for storing the tree)
>
> Anurag Sharma
>
>
>
> On Sun, Jun 13, 2010 at 9:25 PM, Jitendra Kushwaha <
> jitendra.th...@gmail.com> wrote:
>
>> Stack can be sorted in O(n^2).
>>
>> @sankalp:
>>  Stack can always be sorted. Why do you think it cant be in some cases ?
>>
>> One can think like insertion sort
>> algo :
>> 1. for i in (1,n)
>>   2. Pop up the top n-1 element and keep nth element in global variable
>> say "hold"
>>   3. while pushing get the position for "hold" and push it there
>>
>> for loop will take O(n) and step 2 will take take O(n) time. So overall
>> O(n^2) complexity
>> Program can be done with recursion using a variable (hence O(1) space).
>> But it will use system stack :)
>>
>>
>> Any comments OR better solution is welcomed??
>>
>> --
>> Regards
>> Jitendra Kushwaha
>> MNNIT, Allahabad
>>
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