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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>
>
>
> --
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> Saurabh
>
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