Clean, clear, precise! BEGINNER friendly solution!!! Logic clear!
IntuitionTo solve this problem, we need to find the largest potential outlier in an array where:

Total numbers = n


( n - 2 ) elements are special numbers.
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n\nTo solve this problem, we need to find the largest potential outlier in an array where:\n\n- Total numbers = n\n\n- ( n - 2 ) elements are special numbers.\n\n- One element of the remaining two is the "sum of (n - 2) special numbers". (lets call this the sum_element)\n\n- The other remaining element is the outlier, which is distinct from both the sum element and the special numbers.\n\n<u>We know that:</u>\n\ntotal_sum = sum of (n-2) elements  +  sum_element  +  outlier\n\n\n*Rearranging the above elements:*\n\noutlier = total_sum - sum of (n-2) elements  -  sum_element\n\n**Since we know that**\nsum_element = sum of (n-2) elements\n\n**We get the equation as:**\noutlier = total_sum - 2 * sum_element\n\n(We will use this equation in code)\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe question asks for largest outlier element in nums array.\n\nAlso, we know that we will be using this equation:\noutlier = total_sum - 2 * sum_element\n\nSo we need to find total_sum and sum_element to find the outlier\n\n1. To find total_sum, its easy! just sum all the elements of the `nums` array by doing `sum(nums)`\n\n2. We also need to store the frequency of every item in `nums` array in `hash_map` (to make sure `sum_element` and outlier are both distinct)\n\n- `sum_element` is just a single element in `hash_map` (which is the sum of (n-2) elements)\n\n3. We iterate through **every key** in the `hash_map`. **Why?** we are  checking every element because one of them HAS to be the **outlier**\n- By iterating through every element in `hash_map` we find the correct pair of `sum_element` and outlier\n\n4. And **HOW** do we find the outlier? using the equation above `potential_outlier = total_sum - 2 * sum_element`\n(renamed outlier to potential_outlier)\n\n5. Once we find this `potential_outlier`, we are not done yet. We need to make sure that its distinct and that its the LARGEST outlier\n\n6. To make sure its distinct => we just check that `potential_outlier != sum_element or hash_map[sum_element] > 1`\n\n7. Then we just find the `largest_outlier` by doing `max(potential_outlier, largest_outlier)`\n\n\n\n# Complexity\nTime complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n- Calculating the total sum: (O(n)), where (n) is the number of elements in the array.\n- Constructing the hash_map: (O(n)).\n- Iterating over the unique keys in the hash_map: (O(m)), where (m) is the number of unique elements, and (m \\leq n).\n- Therefore, the overall time complexity is (O(n)).\n\nSpace complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity is dominated by the space used for the frequency map, which stores the count of each unique number. Hence, the space complexity is (O(m)), where (m) is the number of unique elements in the array.\n\n# Code\n```python3 []\nclass Solution:\n    def getLargestOutlier(self, nums: List[int]) -> int:\n        \n        #total_sum = sum of (n-2) elements + sum_element + outlier\n\n        #total_sum - sum of (n-2) elements - sum_element = outlier\n\n        # sum_element = sum of (n-2) elements\n        # outlier = total_sum - 2 * sum element\n\n        total_sum = sum(nums)\n\n        hash_map = {}\n        for item in nums:\n            if item in hash_map.keys():\n                hash_map[item] += 1\n            else:\n                hash_map[item] = 1\n        \n\n        largest_outlier = float(\'-inf\')\n\n\n\n        for sum_element in hash_map.keys():\n            potential_outlier = total_sum -  2 * sum_element\n\n            if potential_outlier in hash_map.keys():\n                if potential_outlier != sum_element or hash_map[sum_element] > 1:\n                    largest_outlier = max(potential_outlier, largest_outlier)\n\n        return largest_outlier\n        \n\n```
['Python', 'Python3']