@dave : ya ryt..
@shravan ; my solution works for the case that dave has told.. so at every
step we hav to push min..
On Mon, Sep 5, 2011 at 9:11 PM, Dave <dave_and_da...@juno.com> wrote:
> @Shravan: You at least have to push equal mins on the min stack.
> Otherwise, with the sequence 2, 1, 1, if you push only 2 and 1 onto
> the min stack, then when you pop the first 1 from the data stack and
> pop the 1 from the min stack, the top of the min stack is 2, but the
> minimum in the data stack is 1. So you push 2, 1, 1 onto the min
> stack. When you pop the first 1 from the data stack, you pop the first
> 1 from the min stack, and still show the min = 1.
>
> Don's solution for data sequence 2, 3, 1, 1 would push (2,2) (3,2),
> (1,1), (1,1) onto one stack, whereas you should push 2, 3, 1, 1 onto
> the data stack and 2, 1, 1 onto the min stack, for a saving of one
> stack item. For the sequence 1, 1, 1, 1, 1, 1, 1, you both would have
> the same number of items stacked.
>
> Dave
>
> On Sep 5, 10:09 am, Shravan Kumar <shrava...@gmail.com> wrote:
> > @sandeep
> >
> > You don't need to store duplicate elements in stack2. When you want min
> > return top element. When an element is popped from stack1, pop stack2
> only
> > if it is equal to popped stack1 element.
> >
> > On Mon, Sep 5, 2011 at 8:21 PM, SANDEEP CHUGH <sandeep.aa...@gmail.com
> >wrote:
> >
> >
> >
> > > In my earlier approach , if the element that is stored in "min" got
> popped
> > > out then we have to search entire stack for min.. so i think my
> earlier
> > > approach will not work.. tell me about that??
> >
> > > and i hav another approach.. also tell me about that also?
> >
> > > consider we have to push the elemnts 12 , 3, 15 ,8 ,2, 9
> >
> > > stack 1 stack 2
> >
> > > 12
> >
> > > 12
> >
> > > first 12 came , push in both
> >
> > > 3
> >
> > > 12
> >
> > > 3
> >
> > > 12
> >
> > > next 3 came , push 3 in first stack , and
> >
> > > push minumum ( stack1 top elem , stack2 top elem ) into
> stack2
> > > .. i.e 3 pushed to 2nd stack
> >
> > > 15
> >
> > > 3
> >
> > > 12
> >
> > > 3
> >
> > > 3
> >
> > > 12
> >
> > > now for all the numbers that are coming , push
> > > them into first stack and push min of both stacks top elements into 2nd
> > > stack min (15 , 3) --> 3 pushed to stack2
> >
> > > 8
> >
> > > 15
> >
> > > 3
> >
> > > 12
> >
> > > 3
> >
> > > 3
> >
> > > 3
> >
> > > 12
> >
> > > min (8,3) ---> 3 pushed to stack2
> >
> > > 2
> >
> > > 8
> >
> > > 15
> >
> > > 3
> >
> > > 12
> >
> > > 2
> >
> > > 3
> >
> > > 3
> >
> > > 3
> >
> > > 12
> >
> > > min(2,3) --> 2 pushed to stack2
> >
> > > 9
> >
> > > 2
> >
> > > 8
> >
> > > 15
> >
> > > 3
> >
> > > 12
> >
> > > 2
> >
> > > 2
> >
> > > 3
> >
> > > 3
> >
> > > 3
> >
> > > 12
> >
> > > min(9,2) -- > 2 pushed to stack2
> >
> > > so if at any time we want to print the minimum element , print the
> stack2
> > > top element ,
> >
> > > and if pop() is performed on stack1 , then perform pop() operation on
> > > stack2 also .
> >
> > > after pop () on stack2 , stack2 top still contains the minimum element
> for
> > > the rest of elements ..
> >
> > > On Mon, Sep 5, 2011 at 5:15 PM, kARTHIK R <k4rth...@gmail.com> wrote:
> >
> > >> +1 Don. Good Solution. We can't save space and time at the same time.
> > >> Either use extra space and do operations fast, or use O(1) for min,
> and if
> > >> min is popped, spend O(n) to spot the next min. Depends on the use
> case.
> >
> > >> Karthik R,
> > >> R&D Engineer,
> > >> Tejas Networks.
> >
> > >> On Mon, Sep 5, 2011 at 5:10 PM, bharatkumar bagana <
> > >> bagana.bharatku...@gmail.com> wrote:
> >
> > >>> +1 sandeep
> > >>> @don: u'r sol is correct , but if the number of elements are very
> huge
> > >>> and the updated min is long numbers , then we are storing the min in
> each
> > >>> element ... its waste of memory ...
> > >>> if the # elements are less, then this is good sol..
> >
> > >>> On Mon, Sep 5, 2011 at 1:02 AM, Nitin Garg <
> nitin.garg.i...@gmail.com>wrote:
> >
> > >>>> Don's solution is correct.
> > >>>> At each push() operation, you update the value of min element upto
> that
> > >>>> depth in stack.
> > >>>> Can be illustrated with the following example -
> >
> > >>>> stack = {}
> > >>>> push(2) stack = {(2,2)}
> > >>>> push(3) stack = {(3,2),(2,2)}
> > >>>> push(1) stack = {(1,1),(3,2),(2,2)}
> >
> > >>>> where b in tuple (a,b) represents the min value upto current depth
> in
> > >>>> stack.
> > >>>> pop() and min() are straight forward.
> >
> > >>>> On Mon, Sep 5, 2011 at 12:48 AM, Don <dondod...@gmail.com> wrote:
> >
> > >>>>> Each element in the stack will contain not only its own value, but
> the
> > >>>>> min value at that depth of the stack:
> >
> > >>>>> struct stackItemStruct
> > >>>>> {
> > >>>>> int value;
> > >>>>> int min;
> > >>>>> struct stackItemStruct *next;
> > >>>>> };
> >
> > >>>>> typedef struct stackItemStruct stackItem;
> >
> > >>>>> class stack
> > >>>>> {
> > >>>>> public:
> > >>>>> stack();
> > >>>>> void push(int v);
> > >>>>> int pop();
> > >>>>> int min();
> > >>>>> private:
> > >>>>> stackItem *_stack;
> > >>>>> };
> >
> > >>>>> stack::stack()
> > >>>>> {
> > >>>>> _stack = 0;
> > >>>>> }
> >
> > >>>>> void stack::push(int v)
> > >>>>> {
> > >>>>> stackItem newItem = new stackItem;
> > >>>>> newItem->value = newItem->min = v;
> > >>>>> if (_stack && (_stack->min < v) )
> > >>>>> newItem->min = _stack->min;
> > >>>>> newItem->next = _stack;
> > >>>>> _stack = newItem;
> > >>>>> }
> >
> > >>>>> int stack::pop()
> > >>>>> {
> > >>>>> int result = 0;
> > >>>>> if (_stack)
> > >>>>> {
> > >>>>> result = _stack->val;
> > >>>>> stackItem *tmp = _stack;
> > >>>>> _stack = tmp->next;
> > >>>>> delete tmp;
> > >>>>> }
> > >>>>> return result;
> > >>>>> }
> >
> > >>>>> int stack::min()
> > >>>>> {
> > >>>>> return _stack ? _stack->min : 0;
> > >>>>> }
> >
> > >>>>> On Sep 4, 12:08 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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