@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 <sinv...@gmail.com> 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 <kartik.sac...@gmail.com>wrote:
>
>> +1 @saurabh...:P
>>
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