Simple O(N) Java solution 
IntuitionThe inuition is to try every number as an outlier, and with that thought in mind, do following things -
Understand that for this number to be a valid o
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The inuition is to try every number as an outlier, and with that thought in mind, do following things -
* Understand that for this number to be a valid outlier, the remaining sum should be able to split in two parts, wherein one part is sum of x numbers, and the remaining part is a number.
* sumWithoutOutlier = totalSum - outlier
if(sumWithoutOutlier / 2) == 0, means it can be split, now check if it is present in nums, for which you can add all numbers to a hashMap with their index.
* if(sumWithoutOutlier / 2) is present then this is valid outlier, take care of edge case such that (sumWithoutOutlier / 2) is not the current nums[i] itself, 
Hence get (sumWithoutOutlier / 2)  from hashMap and check if it is not equal to curr idx, if not then it is valid outlier.



# Approach
<!-- Describe your approach to solving the problem. -->

# 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
```java []
class Solution {
    public int getLargestOutlier(int[] nums) {
        int sum = 0;
        Map<Integer, Integer> hMap = new HashMap<Integer, Integer>();

        int outlier = Integer.MIN_VALUE;

        for(int i = 0; i < nums.length; i++){
            sum += nums[i];
            hMap.put(nums[i], i);
        }

        for(int i = 0; i < nums.length; i++){
            int curr = sum - nums[i];
            if(curr % 2 == 0){
                if(hMap.containsKey(curr / 2) && hMap.get(curr / 2) != i){
                    outlier = Math.max(outlier, nums[i]);
                }
            }
        }
        return outlier;
    }
}
```
['Java']