I think it's wrong approach. Can you explain with an example?

On Wed, Jul 27, 2011 at 12:33 PM, pacific :-) <[email protected]> wrote:

> O(nlogn)
> 1. precompute the minimum of [ i+1 ....N] and store in b[i]
> 2. Now do a binary search for a[i] in b[i+1] in the range of  b[i+1..... N]
>
>
> On Wed, Jul 27, 2011 at 12:27 PM, salvador_cerinza <
> [email protected]> wrote:
>
>> 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 <[email protected]>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 <
>>> [email protected]> wrote:
>>>
>>>> i m  suggesting stack  not just for best case only .
>>>>
>>>>
>>>> On Wed, Jul 27, 2011 at 12:16 PM, Pankaj <[email protected]>wrote:
>>>>
>>>>> Even in array best case can be O(n). Why use stack?
>>>>> On Wed, Jul 27, 2011 at 12:14 PM, salvador_cerinza <
>>>>> [email protected]> 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 <
>>>>>> [email protected]> wrote:
>>>>>>
>>>>>>> O(n^2) algo is trivial. Can anybody think of a better approach ???
>>>>>>>
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>
>
>
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> regards,
> chinna.
>
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