wont work for all cases...ignore
i will post the algo....once i fix it
On 25 Mar 2012 17:06, "Amol Sharma" <[email protected]> wrote:

> @atul : it would be better for all to understand if you write the algo
> instead of writing the code..
> --
>
>
> Amol Sharma
> Third Year Student
> Computer Science and Engineering
> MNNIT Allahabad
>  <http://gplus.to/amolsharma99> 
> <http://twitter.com/amolsharma99><http://in.linkedin.com/pub/amol-sharma/21/79b/507><http://www.simplyamol.blogspot.com/>
>
>
>
>
>
>
> On Sun, Mar 25, 2012 at 4:51 PM, atul anand <[email protected]>wrote:
>
>> @shady : yes i guess this is what question says:-
>> so acc to this below algo work , i didnt execute it but i guess it will
>> work
>>
>> void nextSmaller(int arr[],int n)
>> {
>> s1 s;
>> int i,next,ele;
>>
>> s.top=-1;
>> push(&s,0);
>>
>> for(i=1;i<n;i++)
>> {
>> next=arr[i];
>>  if(isEmpty(&s))
>> {
>>       ele=pop(&s);
>>       while(arr[ele] > next)
>>       {
>>  swap(arr,ele,i);
>>                   next=arr[ele];
>> if(isEmpty(&s)==0)
>> {
>> break;
>>  }
>>   ele=pop(&s);
>>       }
>>       if(ele > next)
>>       {
>> push(&s,ele);
>>       }
>>
>> }
>>
>> push(&s,i);
>>  }
>>
>> }
>>
>>
>> On Sun, Mar 25, 2012 at 4:36 PM, shady <[email protected]> wrote:
>>
>>> @gene
>>> i think for  3 4 2 you need to start from left most element, and then
>>> make substitutions one by one.
>>> so it will be
>>> 3 4 2
>>> 2 4 3
>>> 2 3 4
>>>
>>>
>>> @all i googled a bit, and found that O(n) solution is possible for it,
>>> any idea ?
>>>
>>> On Sun, Mar 25, 2012 at 1:59 PM, Kartik Sachan 
>>> <[email protected]>wrote:
>>>
>>>> +1 @saurabh...:P
>>>>
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