Check if Sum of all remaining equal double of one item
# Intuition\nThe value of outlier will be equal to the sum of all the elements subtracted by twice of the actual sum of elements\n\n# Complexity\n- Time complexity:O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass Solution:\n    def getLargestOutlier(self, nums: List[int]) -> int:\n        #nums.sort()\n        freq = Counter(nums)\n        ans = -1001\n        ts = sum(nums)\n        n = len(nums)\n        for i in range(n):\n            outlier = ts - (2*nums[i])\n            #print()\n            if freq[outlier] - (1 if outlier == nums[i] else 0):\n                ans = max(ans,outlier)\n        return ans\n\n\n```
['Python3']