Sorry but I have still not got the solution u have tried to propose here.
Firstly the space complexity remain O(n) only what I said.

To understand thing u said better lets take an example of stack with
following entries

--------------------------->top
45  22  56 44 55 3  2  4 -999 4 2  45 66 76 44343

how will your second stack look like and how will the push/pop/min/max will
work here ?


On Thu, Oct 7, 2010 at 11:33 PM, Saurabh Gupta
<[email protected]>wrote:

>
>
> On Tue, Oct 5, 2010 at 9:47 AM, saurabh singh <[email protected]>wrote:
>
>> elaborate plz
>
>
> For every new element in stack, you need thrice of space to store the min
> and max elements also. This has the effect that at state of stack, you can
> get the min and max till that point. Instead of this, maintaining a new
> stack for min elements would be much more efficient in terms of memory since
> in that all the number of elements would be lesser than the main one.
>
> so, basically in your solution, the size of object will be three times
> bigger than the data type which can hold the number otherwise.
>
>
>>
>> On Tue, Oct 5, 2010 at 9:42 AM, Saurabh Gupta <[email protected]
>> > wrote:
>>
>>> In this method, the extra memory is used. In fact, maintaining a separate
>>> stack of min and max would consume lesser memory than this.
>>>
>>> On Thu, Sep 30, 2010 at 12:17 AM, saurabh singh 
>>> <[email protected]>wrote:
>>>
>>>> You will just need to see what is min and max available on the current
>>>> top before push. in case of pop u dnt need to do anything...
>>>>
>>>> consider this array imp of stack
>>>> each array index is stored with this object : {data, min_till_here,
>>>> max_till_here}
>>>>
>>>> ------------------------------------------------------------->Top
>>>> [{4,4,4} , {2,2,4}, {7,2,7} , {-6,-6,7}, {1,-6,7}, {9,-6,9}]
>>>>
>>>>
>>>>
>>>> So if u push say 20 then at top u see whats max and min till now. since
>>>> curr min (-6) is smaller than 20 so min remains unaffected and since curr
>>>> max (9) is smaller than 20 so curr max becomes 20. Hence the object at top
>>>> in stack looks like {20,-6,20}
>>>>
>>>>
>>>> On Wed, Sep 29, 2010 at 10:18 PM, albert theboss <
>>>> [email protected]> wrote:
>>>>
>>>>>
>>>>> when we pop out something ....
>>>>> we need to find the max min again if max or min is popped out...
>>>>> this ll take again O(n) to find max and min....
>>>>>
>>>>> Correct me if am wrong....
>>>>>
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>>>>
>>>>
>>>> --
>>>> Thanks & Regards,
>>>> Saurabh
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Saurabh

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