suppose in that max stack max elements are maitained.suppose one element
with value less arrives.u need to insert it in proper pos.how is that
possible in 0(1) ?

On Sun, Sep 4, 2011 at 11:15 PM, Deepak Garg <deepakgarg...@gmail.com>wrote:

> maintain a separate stack containing min and max element at each step.....
> so if u pop an element for the original stack, pop from the second stack
> also...
>
>
>
> On Sun, Sep 4, 2011 at 11:13 PM, Deepak Garg <deepakgarg...@gmail.com>wrote:
>
>> +1 sandeep
>>
>> On Sun, Sep 4, 2011 at 11:11 PM, SANDEEP CHUGH 
>> <sandeep.aa...@gmail.com>wrote:
>>
>>>
>>> we can take another variable min..
>>>
>>> first time push operation is done , store the element into  min..
>>>
>>> next time push is performed , compare the number u r pushing with the
>>> already stored no in "min"  variable.. and store minimum of two no's in min
>>> variable.. and thn perform the push operation..
>>>
>>> so min always contain the minimum no...
>>>
>>> tell me will that approach work or not?
>>>
>>> On Sun, Sep 4, 2011 at 10:38 PM, Sangeeta <sangeeta15...@gmail.com>wrote:
>>>
>>>> How would you design a stack which,in addition to push and pop,also
>>>> has a function min which returns the minimum element?push,pop and min
>>>> should all operate in O(1) time
>>>>
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