elaborate plz

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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-- 
Thanks & Regards,
Saurabh

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