@apoorve

say the input is 9 8..
I think it will fail...

Secondly, i see that max is not being used....


On Dec 31, 8:19 pm, Apoorve Mohan <[email protected]> wrote:
> Hey
>
> Below is a solution. Hope it works.
>
> Let 'n' be the size of the array we have.
>
> max = min = array(n-1)
> flag=1;
>
> for(i=(n-2) to 0)
> {
>   if(array(i)<array(i+1))
>   {
>      if(min < array(i))
>      {
>         flag = 0;
>         break;
>      }
>      else
>      {
>         min = array(i);
>      }
>   }
>   else
>   {
>      if(min > array(i))
>      {
>         flag = 0;
>         break;
>      }
>      else
>      {
>         max = array(i);
>      }
>   }
>
> }
>
> flag?printf("yes"):printf("no");
>
>
>
>
>
>
>
>
>
> On Fri, Dec 30, 2011 at 1:21 PM, top coder <[email protected]> wrote:
> > Input : You have been given a sequence of integers.
> >  Now without actually constructing BST from the given sequence of
> > integers (assuming the sequence is pre-order) determine if each node
> > of BST has single child (either left or right but not both).
> >  Output : YES if given integer sequence corresponds to such a BST.
> > otherwise say NO.
>
> > Ex. Say NO for 16,10,8,9,29,27,22.
> >  Say YES for 10,19,17,14,15,16
>
> > I know the algo O(N^2) as follows:
> > For every node in array(traverse from left to right),
> >  all the other nodes must be less or all the nodes must be greater
> > than it.
>
> > But interviewer is expecting O(N). Any ideas?
>
> > --
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>
> --
> regards
>
> Apoorve Mohan

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