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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