@ 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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