suppose k is the sum to be found.
@vaibhav: yes it will stop when crossing is there means (j must be greater
than i).initially sum = a[i] + a[j] (where i = 0 n j = n- 1) n we will
increase i when sum is less than k and decrease j when sum < k.stop if sum
== k. and  if no i , j found till j > i. then pairs not possible,

On Mon, Jun 27, 2011 at 8:28 AM, Bharath Soma <[email protected]>wrote:

> @ankit, sorry i was mistaken its O(nlogn) for searching the two elements...
>
>
> On Mon, Jun 27, 2011 at 8:04 PM, Swathi <[email protected]> wrote:
>
>> Dave,
>>
>> Can you provide the psuedo code for this..
>>
>> Thanks,
>> Swathi
>>
>>
>> On Mon, Jun 27, 2011 at 7:30 PM, Dave <[email protected]> wrote:
>>
>>> @Sunny. Mea culpa. You are correct. Revised (and correct) algorithm.
>>> Do two inorder traversals, one in the usual (descend to the left
>>> before descendung to the right) direction and the other in the
>>> reversed(descend to the right before descending to the left)
>>> direction. Let u and r be the current nodee of the two traversals,
>>> respectively. If u + r < x, then advance the usual traversal and
>>> repeat the comparison. If u + r > x, advance the reverse traversal and
>>> repeat the comparison. If u + r = x, and if u != r, then terminate
>>> with success. If u = r, then terminate with failure.
>>>
>>> Dave
>>>
>>> On Jun 27, 7:53 am, sunny agrawal <[email protected]> wrote:
>>> > @Dave
>>> > i think your solution won't work
>>> > consider inorder traversal of a BST is 1 6 7 8 15 and x = 14
>>> > initially both u,v (1,1)
>>> > according to u your algorithm will proceed like
>>> > (1,1) -> (1,6) -> (1,7) -> (1,8) -> (1,15) -> (6,15) ............ ->
>>> (15,15)
>>> >
>>> > and clearly in second step of your solution if (u+v) > x after
>>> advancing u
>>> > still u+v will be greater than x
>>> > so something is wrong
>>> > I think your solution will work in case we need to find 2 nodes with
>>> > difference x.
>>> >
>>> > correct me if i am wrong.!!
>>> >
>>> >
>>> >
>>> >
>>> >
>>> > On Mon, Jun 27, 2011 at 6:13 PM, Dave <[email protected]> wrote:
>>> > > @Nishant: No need to store the data in an array. Do two inorder
>>> > > traversals simultaneously. Let u and v be the current nodes of the
>>> two
>>> > > traversals, respectively. If u + v < x, then advance the "v"
>>> > > traversal. If u + v > x, advance the "u" traversal.
>>> >
>>> > > Dave
>>> >
>>> > > On Jun 27, 3:40 am, Nishant Mittal <[email protected]>
>>> wrote:
>>> > > > do inorder traversal of tree and store values in an array.
>>> > > > Now find pairs by applying binary search on array..
>>> >
>>> > > > On 6/27/11, manish kapur <[email protected]> wrote:
>>> >
>>> > > > > --
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>
>
>
> --
> ...Thanks
> Bharath
>
>
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-- 
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Indian Institute of Information Technology Allahabad.
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