(for array sorted in ascending order)
take 2 indexes i and j pointing to 1st and last element of the array
respectively...
now if(arr[i]+arr[j] == x)
print(arr[i]);
print(arr[j];
else if(arr[i]+arr[j]>x)
j--;
else
i++;

I think this should work...(i've not checked)
correct me if i m wrong


On 6/27/11, ankit sambyal <ankitsamb...@gmail.com> wrote:
> @Bharath : Cud u plz explain how r u searching the elements in O(n) time?
> Because if we use binary search, it will have O(n*log n )  worst case
> time complexity. One way in which I think it cud be made O(n) is that
> we can use a hash table, with a good hash function apart frm the
> array. And then for each element 'm' in the array, we cud search if
> there is an element (sum - m) in O(1) time by using hash table. Still
> we can't assure O(n) time complexity. Because coming up with a good
> hash function is not easy. Again, hash table takes more space....
>
>
>
>
> On Mon, Jun 27, 2011 at 1:40 AM, Nishant Mittal
> <mittal.nishan...@gmail.com> 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 <manishkapur.n...@gmail.com> wrote:
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