@topcoder..
You are correct.. I just figured out that the N<=3 condition is not
always correct..
Below i have explained with example the code that will fix it:
Basically for N=3 , say the sequence is a , b, c..
Then the valid cases would be:
a < b < c
a >= b >= c
a < b , (b>=c and a < c)
a >= b , ( b < c and c <= a)
as u can see.. for 4, 2, 6
4 > 2 and ( 2 < 6 but 6 > 2) hence its invalid..
if ( A[0] < A[1] )
{
if ( (A[1] < A[2]) )
{
left = A[0];
mid = A[1];
right= A[2];
}
else if (A[1] >= A[2] && A[0] < A[2] )
{
left = A[0];
mid = A[2];
right = A[1];
}
else
goto end; printf("NO");
}
else
{
if (A[1] >= A[2])
{
left = A[2];
mid = A[1];
right = A[0];
}
else if ( (A[1] < A[2]) && (A[0] >= A[2]) )
{
left = A[1];
mid = A[2];
right = A[0];
}
else
goto end; printf("No");
}
// Replace the greatest, smallest and mid code with the above one and
it shall work..
On 30 Dec, 13:42, top coder <topcode...@gmail.com> wrote:
> @Lucifer
>
> Looks like Your algo works for all of the my test cases. I am not able
> to find counter case.
>
> Also we should not Print "Yes" for N<=3
>
> For eg:
>
> A = 4,2,6
>
> Then 2 is a root with 4 as left child and 6 as right child. It
> contains two childs and fails the case.
>
> On Dec 30, 1:26 pm, Lucifer <sourabhd2...@gmail.com> wrote:
>
>
>
>
>
>
>
> > @above
>
> > Also i would like to mention that the above algo considers equal
> > valued elements and assumes that the BST should have the following
> > property:
>
> > 1) leftChild < = root
> > 2) rightChild > root
>
> > On 30 Dec, 13:24, 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?- Hide quoted text -
>
> > - Show quoted text -
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