Easiest solution:

• Put all elements in a hash map -> O(nlogn)
• Iterate through list doing a lookup for (Sum - A[i]) -> O(n) with O(1) lookups
• If Sum is 2x A[i], check for second instance value in the map

Solution in O(nlogn)

Easiest solution:

• Put all elements in a hash map -> O(n) to O(n^2) depending on has function
• Iterate through list doing a lookup for (Sum - A[i]) -> O(n) with O(1) lookups
• If Sum is 2x A[i], check for second instance value in the map

Solution in O(n) to O(n^2) depending on has function