yes i too think now that it should work..but on every push/pop we will need
to update the other two stacks also which can be done in constant time..

On Fri, Oct 8, 2010 at 12:58 AM, tech rascal <[email protected]>wrote:

> I think saurabh gupta is rite.....if v take 2 extra stacks ...1 for min and
> 1 for max, thn some space wud b saved.
> for the above example .........max_stack wud b-
>
> ------------------------>top
> 45 56 66 76 44343
>
> and min_stack wud b-
>
> --------------->top
> 45 22 3 2 -999
>
> so, here v need 2 save only 5 elements in max_stack, 5 elements in
> min_stack and 15 elements in full_stack ( acc 2 above example only), hence
> total=25 elements..........othrwise if u do it by taking only 1 stack thn u
> need space for ..15X3 (45)elements.
>
> tell me if I am wrong..
>
> On Thu, Oct 7, 2010 at 11:49 PM, saurabh singh <[email protected]>wrote:
>
>> 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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>>
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>> Saurabh
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Thanks & Regards,
Saurabh

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