@topcoder..
As pointed u by earlier that 4 ,2, 6 doesn't work for the initial
code..
The 1st solution that i gave to resolve it is still faulty.. it will
fail for 4 3 2 1.. basically increasing or decreasing set of nos..
// The second and 3rd approach would work fine.. i.e (-INF, +INF) and
(smallest -1 , greatest +1)
The 1st approach of initialization fails because we are restricting
the max and min value that can held by the BST without knowing their
exact values.. Hence, for a completely sorted sequence of nos.(inc or
dec) it will fail....
I m putting the final code below to remove all confusion :
------------------------------------------
int left = -INF; // or scan the array to get the smallest element
//and assign left to (smallest -1) assuming that
//smallest is not the min value that an integer can
hold.
int right = +INF; // or scan the array to get the greatest element
//and assign left to (greatest +1) assuming
that
//greatest is not the max value that an integer
can hold.
int mid = A[0];
int i = 0;
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;
}
if(i == N)
printf("YES");
else
printf("NO");
On 31 Dec, 10:19, Lucifer <sourabhd2...@gmail.com> wrote:
> @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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