Let say stack S.
1.insert elements in S of A[] from right to left.
2.int val = S.top();
3.S.pop();
4.now check val with S.top() until u find any element smaller than val.
5.Note down the element pop it from stack
6.if step 4 is true , the push val in stack S and all elements which were
popped in the order they were popped except the last matched candidate
element.

Yeah..dis algo is not very efficient..

On Wed, Jul 27, 2011 at 12:20 PM, Pankaj <jatka.oppimi...@gmail.com> wrote:

> Can you please elaborate a little about your stack based solution. I was
> thinking of using queue but was unable to make a perfect algo.
>
>
> On Wed, Jul 27, 2011 at 12:18 PM, salvador_cerinza <
> vishwakarma.ii...@gmail.com> wrote:
>
>> i m  suggesting stack  not just for best case only .
>>
>>
>> On Wed, Jul 27, 2011 at 12:16 PM, Pankaj <jatka.oppimi...@gmail.com>wrote:
>>
>>> Even in array best case can be O(n). Why use stack?
>>> On Wed, Jul 27, 2011 at 12:14 PM, salvador_cerinza <
>>> vishwakarma.ii...@gmail.com> wrote:
>>>
>>>> Best case : O(n)
>>>> Worst case : O(n^2)
>>>> can be done using stack.
>>>>
>>>> Thinking of better solution. .
>>>>
>>>>
>>>> On Wed, Jul 27, 2011 at 11:50 AM, ankit sambyal <ankitsamb...@gmail.com
>>>> > wrote:
>>>>
>>>>> O(n^2) algo is trivial. Can anybody think of a better approach ???
>>>>>
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