Solution using Python, hash map. Not the fastest, but intuitive.
# Intuition\nIf we consider a value as an outlier, then there\'s another value, which will be the sum of the rest of the values. So, that value will basically be the half of the sum of the remaining values. Therefore, we can just sum the values other than the outliers, divide by 2, and check if that value exists in the subarray.\n\n# Approach\nSum all the values. Consider each value n in nums as an outlier. Get sum of all the other values. Make sure the new sum is even, which means it\'s divisible by 2. Divide that new sum by 2. If that sum exists in the remaining array, that\'s the outlier.\n\nUse a frequency hashmap to check if a number exists in the subarray\n\n# Complexity\n- Time complexity:\n$$O(n)$$ - Each of the sum operation, freq hash creation operation, and looping through the nums array takes O(n) time. The operations inside the loop takes constant time\n\n- Space complexity:\n$$O(n)$$: The freq hash takes O(n) space for worst case.\n\n# Code\n```python3 []\nclass Solution:\n    def getLargestOutlier(self, nums: List[int]) -> int:\n        # sum all the values\n        # consider a num as outlier. Get sum of all the other values\n        # Divide that new sum. If that sum exists in the remaining array,\n        # that\'s the outlier\n        # Use a frequency hashmap to check if a number exists in the subarray\n\n        # sum all the values\n        sumVals = sum(nums)\n        # frequency hashmap\n        freqHash = Counter(nums)\n        # track largest outlier\n        maxO = float("-inf")\n\n        for n in nums:\n            # adjust occurance of n in freq hash\n            if n in freqHash and freqHash[n] > 0:\n                freqHash[n] -= 1\n            # get sum of all other vals, check if even, and get divided val\n            sumOtherVals = sumVals - n\n            if sumOtherVals % 2 == 0:                \n                dividedVal = sumOtherVals / 2\n                if dividedVal in freqHash and freqHash[dividedVal] > 0:\n                    maxO = max(maxO, n)\n            if n in freqHash and freqHash[n] >= 0:\n                    freqHash[n] += 1\n        \n        return maxO\n```
['Hash Table', 'Python3']