simple approach  100% beats || TC->O(NLOGN) || SC->O(N) || JUST ONE FOR LOOP|| C++||
IntuitionApproachComplexity
Time complexity:

Space complexity:
Code
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->

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

# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->

# Code
```cpp []
class Solution {
public:
    double minimumAverage(vector<int>& nums) {
        int n = nums.size();
        vector<double>ans;
        sort(nums.begin() , nums.end()); //sortthe array  
        for(int i =0 ;i<n/2 ;i++){
              ans.push_back((nums[i] + nums[n-i-1]) / 2.0); //find everytime avarage
        }


        return *min_element(ans.begin() , ans.end()); // just return the small avarage  
        
    }
};
```
['C++']