while ( ++i < N)
{
if (A[i] <= mid && A[i] > left)
{
*mid = A[i]; // both mid and right will contain same value**
right = mid;*
// right=mid; /* i guess this is what you want....*/
// mid=A[i]
*
*
}
else if (A[i] > mid && A[i] <= right)
{
*mid = A[i]; // * *// both mid and left will contain same value*
*
left = mid;*
}
else
break;
}
bcozz every node should have single child then how should
i interpret mid,left,right ??
On Fri, Dec 30, 2011 at 1:54 PM, Lucifer <sourabhd2...@gmail.com> wrote:
> @An idea...
>
> Let the array of integers be A[N]..
>
> If N < = 3, then the answer would be YES.
>
> If N > 3, then the following algo will follow:
> // The algo is self explanatory as its just ensuring BST property
> keeping in mind that a parent node has only one child...
>
>
> int left = smallest(A[0], A[1], A[2]) ;
> int right = greatest(A[0], A[1], A[2] ;
> int mid = A[0] + A[1] + A[2] - left - right;
>
> int i = 2;
>
> while ( ++i < N)
> {
> if (A[i] <= mid && A[i] > left)
> {
> mid = A[i];
> right = mid;
> }
> else if (A[i] > mid && A[i] <= right)
> {
> mid = A[i];
> left = mid;
> }
> else
> break;
> }
>
> if ( i == N)
> printf ("YES");
> else
> print("NO");
>
>
> On 30 Dec, 12:51, top coder <topcode...@gmail.com> 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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