Easy Python O(N) solution
IntuitionFind the criteria to decide if each number can be a outlierApproachCalculate the sum of the array
For each number in the array, if the number is the ou
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Find the criteria to decide if each number can be a outlier
# Approach
<!-- Describe your approach to solving the problem. -->
Calculate the sum of the array
For each number in the array, if the number is the outlier, the remaining sum / 2, which should be another special number should still be in the remaining array.

Use hashmap to record the nums frequency. so that we can just check if the count[potential_sum] value
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
O(N)
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
O(N)
# Code
```python3 []
class Solution:
    def getLargestOutlier(self, nums: List[int]) -> int:
        total = sum(nums)
        count = Counter(nums)
        result = -1001

        for num in nums:
            # in case the potential_sum == num
            count[num] -= 1
            # remaining sum
            curr_sum = total - num
            # test if this number can be a outlier
            potential_sum = curr_sum / 2
            if potential_sum in count and count[potential_sum] > 0:
                result = max(result, num)
            count[num] += 1       
        return result
```
['Python3']