On Tue, Oct 5, 2010 at 9:47 AM, saurabh singh <saurabh.n...@gmail.com>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
> <saurabhgupta1...@gmail.com>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
>> <saurabh.n...@gmail.com>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 <
>>> alberttheb...@gmail.com> 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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> Thanks & Regards,
> Saurabh
>
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