That's not really elogant...

By k, I assume you are talking about k=alen(arr) ?

You do find the maxium value in log time.

But I think it might be better, just to:

* first we find the max element
max_element = 0
max_index = 0
for k = alen(arr) to 1 step -1
if X >= arr(k)
max_element = arr(k)
max_index = k
endif
endfor

*This solution is much easier if I was programming in c/c++ where I could *use pointers. We'll wing it tho ;) Really, what I want to do is have a *pointer to the max_element and one to the min_element, and then if their *sum is greater then x, I decrease the max_element, if their sum is less *then X I increase the min_element, if their sum equal X I end. One final
*case, our end case, if max_element = min_element and max_ele + min_ele != *X, I return failure.

max_index = k
min_index = 1
do while arr(max_index)+arr(min_index) != X
if arr(max_index)+arr(min_index) > X
max_index = max_index - 1
else if arr(max_index)+arr(min_index) < X
min_index = min_index + 1
else if arr(max_index) = arr(min_index)&&we already know their sum doesn't
&&equal X, that's tested by our * &&our while loop
return "Failure!"
endif
enddo

Worst case, the entire array is scanned, or O(n) - linear running time.

im almost absolutely sure this works,

matt

>test = arr[int(k/2)]
>if X < test && We don't need to check the rest of the array
> test = arr[int(k/4))
> if X < test && we don't need to check 3/4 of the array
> ...
> endif
>endif