Mike,

>>>>>>>>>>
*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,
>>>>>>>>>>

Bingo! Your solution is correct with one exception. I did not specify the array must contain two numbers that sum to X. Changing your do while to include an additional condition to check min_index < max_index - and returning a no match exists if ever false - solves the problem.

The proof is a little bit tricky because it involves proving that any rejected by the algorithm cannot be part of the solution.

Congratulations!

Daniel