@varun: Thanks...

On Mon, Jun 27, 2011 at 9:18 PM, varun pahwa <[email protected]>wrote:

> 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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>
>
>
> --
> Varun Pahwa
> B.Tech (IT)
> 7th Sem.
> Indian Institute of Information Technology Allahabad.
> Ph : 09793899112 ,08011820777
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>
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