@topcoder
1) For N <= 2 and not 3 its always a "YES"..
2) Also if u are OK with traversing the array twice i.e instead of a
single run N we can do it 2*N..
Then we can do the following to intialize left and right properly
instead of -+INF :
1) Scan the array to get the smallest and the greatest element..
2) Assign left = (smallest - 1) and right = (greatest + 1), given
the fact that smallest and greatest are not already at the integral
boundaries..
On 30 Dec, 14:19, Lucifer <sourabhd2...@gmail.com> wrote:
> Actually, the initial code that i have written could have written as
> follows:
>
> int left = -INF; // or smallest integral value
> int mid = 10; // greatest integral value
> int right = +INF;
>
> int i = 0;
> while (++i < N)
> {
> // Same as the previous code..
>
> }
>
> I didn't take the above initialization approach as while writing
> actual code i personally don't find -INF and +INF to be a good option
> to initialize.. Also instead of -INF and +INF we could have replaced
> it with smallest integral and greatest integral value but still I
> don't prefer it until and unless the node values are not guaranteed to
> be within a particular range...
>
> ---------------------------
>
> On 30 Dec, 14:14, Lucifer <sourabhd2...@gmail.com> wrote:
>
>
>
>
>
>
>
> > @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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