# Naive approach - O(n)
def twoNumberSum(array, targetSum):
    for num in array:
        if num == 0:
            return sorted([num, targetSum])
        if (num < targetSum):
            diff = targetSum - num
            if diff in array:
                if array.index(diff) != array.index(num):
                    result = [num, diff]
                    return sorted(result)

    return []

# Better approach - O(nlog(n))
def twoNumberSum(array, targetSum):
    array.sort()
    left_pointer = 0
    right_pointer = len(array) - 1
    while left < right: # While the pointers don't overlap...
        currentSum = array[left_pointer] + array[right_pointer]
    if currentSum == targetSum:
        return [array[left_pointer], array[right_pointer]]
    # Increase or decrease based on size (list is already sorted)
    elif currentSum < targetSum:
        left_pointer += 1
    elif currentSum > targetSum:
        right_pointer -= 1
    return []