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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