why to bother this much...? just count the elements when popping and
output the middle one .
while(!s.empty()){
e= s.pop()
count++
q.enq(e);
}
count <<= 2;
while(count){
e = q.deq();
s.push(e);
count --;
}
output s.top()
while(!q.empty()){
e = q.deq();
s.push(e);
}
On Aug 22, 4:27 pm, Shravan Kumar <shrava...@gmail.com> wrote:
> Pop each element and en-queue it twice and de-queue it once. When stack is
> empty the front of the queue will be middle element.
>
>
>
>
>
>
>
> On Mon, Aug 22, 2011 at 4:01 PM, Ankur Garg <ankurga...@gmail.com> wrote:
> > Find the middle of the stack..(Time complexity should be minimum)
>
> > Stack is not implemented as Linked List ...u have normal stack with
> > push,pop and top
>
> > How to do this ??
>
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