single traversal + unordered_map stl => O(n) 
IntuitionApproach[a,b,c,d,e] suppose b is an outlier, for each i we will be checking whether (a+b+c+d+e) - (a+c+d+i) is present in our array or not and handling
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->

# Approach
[a,b,c,d,e] suppose b is an outlier, for each i we will be checking whether (a+b+c+d+e) - (a+c+d+i) is present in our array or not and handling some edge-cases, if present then that is our outlier and i will be our special number.

# Complexity
- Time complexity:  O(n)
- Space complexity: O(n)

# Code
```cpp []
class Solution {
public:
    int getLargestOutlier(vector<int>& nums) {

        int n = nums.size() ;
        int sum = accumulate(nums.begin(), nums.end(), 0) ;
        
        unordered_map<int,int> mp ;
        for(auto i : nums)  mp[i]++; 
      
        int ans = -9000 ;

        for(int i=0;i<n;i++){
            int x = sum - (nums[i] + nums[i]) ;
            
            if(mp.find(x)!=mp.end()) {
                if(x==nums[i] and mp[x]==1) continue ;
                ans=max(ans, x);
            }
        }
        if(ans==-9000) return -1 ;
        return ans;
    }
};
```
['Hash Table', 'C++']