only 2 line answer easy to understand both line explain good
Approach\n Describe your approach to solving the problem. \nwe have 3 case for 2 element remove\n1. 2 minimum\n2. 2 maximum\n3. 1 minimum and 1 maximum\n\nfor a
# Approach\n<!-- Describe your approach to solving the problem. -->\nwe have 3 case for 2 element remove\n1. 2 minimum\n2. 2 maximum\n3. 1 minimum and 1 maximum\n\nfor all cases low always 0 \nand high is (as a que maximum-minimum) after remove \n2 element \n\n1. ```nums[nums.size()-1]-nums[2]```\n2. ```nums[nums.size()-3]-nums[0]```\n3. ```nums[nums.size()-2]-nums[1]```\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n    O(nlogn)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->  \n    O(1)\n# Code\n```\nclass Solution {\npublic:\n    int minimizeSum(vector<int>& nums) {\n        \n        sort(nums.begin(),nums.end());\n       \n       return min({nums[nums.size()-2]-nums[1],nums[nums.size()-1]-nums[2],nums[nums.size()-3]-nums[0]});\n       \n    }\n};\n```
['C++']