@Prem: The approach has been discussed in previous posts in this
thread, so there is no need for Shravan to repeat it. Please follow
your own advice and tell what problem shows up with the sample input.
Dave
On Sep 6, 8:26 am, Prem Krishna Chettri <hprem...@gmail.com> wrote:
> @ Shravan :- Thank for the Code but there are couple of issue with that..
>
> 1> Please try to provide the algo approach. Coz I feel
> here we discuss irrespective of language , the technique of approach.
> 2> Your Code implementation of whatever your approach
> is WRONG..
>
> Please Run through these sample Inputs
> Values :- 233556263
>
> You need to make grow both the
> stack Simultaneously.
>
> Well Guys the Algo I presented above is my understanding as I feel its
> simple topic being highly exaggerated. Please feel free to break it if I
> have missed something..
>
>
>
> On Tue, Sep 6, 2011 at 6:26 PM, Shravan Kumar <shrava...@gmail.com> wrote:
> > void push(int num){
> > if(stk1.isEmpty()){
> > stk1.push(num);
> > stk2.push(num);
> > }
> > else{
> > if(num>stk2.top())stk1.push(num);
> > else{stk1.push(num);stk2.push(num)}
> > }
> > }
>
> > int pop(){
> > int num=stk1.pop();
> > if(num==stk2.top())stk2.pop();
> > return num;
> > }
>
> > On Tue, Sep 6, 2011 at 6:12 PM, Prem Krishna Chettri
> > <hprem...@gmail.com>wrote:
>
> >> Guys What the Issue Here?? I think its straight forward.
>
> >> If I hv two Stack
> >> First :- Keep pushing and Popping the incoming values
> >> Second :- Keeping track of the so far min element in the First Stack.
>
> >> Now maintaining second stack is bit tricky.
> >> PUSH :- If the element is first element Push the Same
> >> element in Second Stack what we Pushed in Stack 1.
> >> Else compare the last Second.top with Current
> >> Element
> >> Push which ever
> >> is smaller (or Equal).
>
> >> POP :- POP both the stack simultaneously.
>
> >> On Tue, Sep 6, 2011 at 5:52 PM, Don <dondod...@gmail.com> wrote:
>
> >>> @HARISH:
> >>> Push 20 1 3 1 5 1 6 1 2 1
> >>> Pop
>
> >>> Now in your algorithm min will return 20, even though 1 and 3 and
> >>> other smaller numbers are still in the stack.
>
> >>> Don
>
> >>> On Sep 6, 6:25 am, "HARISH S.C" <s.c.har...@gmail.com> wrote:
> >>> > Have a separate stack for minimum. While pushing, insert the number in
> >>> > minimum stack only if the given number is less that or equal to the
> >>> number @
> >>> > the top of min stack. While removing, remove the value from min stack
> >>> only
> >>> > if its equal to the value thats popped.
>
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