Easiest Solution Available
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n- O(nlogn)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n- O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    double minimumAverage(vector<int>& nums) {\n\n        vector<long double> arr;\n        sort(nums.begin(),nums.end());\n        int start=0;\n        int end=nums.size()-1;\n\n        while(start<end){            \n            long double ans=(nums[start]+nums[end])/2.0;\n            arr.push_back(ans);\n            start++;\n            end--;\n        }\n\n        sort(arr.begin(),arr.end());\n        return arr[0];\n    }\n};\n```
['C++']