@above format got messed up and re-posting..
while ( ++i < N)
{
if (A[i] <= mid && A[i] > left)
{
right = mid;
mid = A[i];
}
else if (A[i] > mid && A[i] <= right)
{
left = mid;
mid = A[i];
}
else
break;
}
On 31 Dec, 10:17, Lucifer <sourabhd2...@gmail.com> wrote:
> @atul.. thanks for pointing out the editing mistake..
>
> I have fixed the while loop below:
>
> while ( ++i < N){ if (A[i] <= mid && A[i] > left) { right =
> mid;
> mid = A[i]; } else if (A[i] > mid && A[i] <= right) {
> left = mid;
> mid = A[i]; } else break;}
>
> On 31 Dec, 01:45, atul anand <atul.87fri...@gmail.com> wrote:
>
>
>
>
>
>
>
> > 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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