Simple Python, O(n) solution (99%) (Simply check if array has half of the Sum of remaining items)
Simply check if half of the sum of remaining items is present in the listComplexity
Time complexity: O(n)

Space complexity: O(n)
Code
Simply check if half of the sum of remaining items is present in the list

# Complexity
- Time complexity: O(n)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity: O(n)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->

# Code
```python3 []
class Solution:
    def getLargestOutlier(self, nums: List[int]) -> int:
        count = Counter(nums)
        total = sum(nums)
        pot_outlier = float(-inf)
        for key in count:
            if key > pot_outlier:
                currtotal = total - key
                count[key]-=1
                if currtotal % 2 == 0 and (currtotal//2 in count) and (count[currtotal//2]>0):
                    pot_outlier = key
                count[key]+=1
        return pot_outlier
```
['Python3']