OK...
suppose our tree is
5
/\
4 6
/ \
3 7
/ \
2 8
9 1 is analogous to 1 9...And the question requires only two nodes,it does
not says about all such pairs.
On Sun, Jul 17, 2011 at 2:52 PM, sagar pareek sagarpar...@gmail.com wrote:
OK...
suppose our tree is
5
/\
OK
On Sun, Jul 17, 2011 at 2:59 PM, saurabh singh saurab...@gmail.com wrote:
9 1 is analogous to 1 9...And the question requires only two nodes,it does
not says about all such pairs.
On Sun, Jul 17, 2011 at 2:52 PM, sagar pareek sagarpar...@gmail.comwrote:
OK...
suppose our tree is
@dave
what if the k= root-left + right most leaf ?
how ur algo works on it?
On Sat, Jul 16, 2011 at 9:28 PM, Dave dave_and_da...@juno.com wrote:
@Noobcoder: To do it in O(n) without destroying the BST, do two
inorder traversals simultaneously, one in the normal forward (left
subtree first)
ok i got it
actually u written wrong that f/w and reverse traversal are running
parallel
u must wrote that f/w traversal inside reverse or vice versa
On Sat, Jul 16, 2011 at 10:35 PM, Dave dave_and_da...@juno.com wrote:
@Sagar: No problem. The algorithm would do only forward traversal
and so it must not be O(n)
On Sat, Jul 16, 2011 at 10:54 PM, sagar pareek sagarpar...@gmail.comwrote:
ok i got it
actually u written wrong that f/w and reverse traversal are running
parallel
u must wrote that f/w traversal inside reverse or vice versa
On Sat, Jul 16, 2011 at 10:35 PM,
Ok may be i m not getting ur logic...
On Sat, Jul 16, 2011 at 11:03 PM, Dave dave_and_da...@juno.com wrote:
@Sagar: No. I didn't say that they were in parallel or that one was
inside the other. Go back and read it again and you will see that I
said that they were being performed
You must take an example and then explain
On Sat, Jul 16, 2011 at 11:59 PM, Dave dave_and_da...@juno.com wrote:
@Sagar: If you are not getting my logic, ask a question.
Dave
On Jul 16, 12:35 pm, sagar pareek sagarpar...@gmail.com wrote:
Ok may be i m not getting ur logic...
On
Use O(2n) memory , list the in-order traversal of BST say A[0..n].
and K-A[0...n] say B . Now apply standard merge function(Merge sort)
on A and B. keeping track of equal found elements during comparison to
get the ans.
On 7/17/11, sagar pareek sagarpar...@gmail.com wrote:
You must take an
@sagar This is what Dave is suggesting in a more pseudocode way:-
1-Traverse a pointer right down to the leftmost element,i.e.the
shortest,say small
2-traverse a pointer left down to the rightmost element i.e.the
largest.say
large
while(small!=large)
3-Compare their sum.If sumk set large to its
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