Simplest Java solution with Time and Space complexity explaination
Approach\n Describe your approach to solving the problem. \nThe code first checks if the length of the input array is 3, and if so, returns 0. This is because, 
# Approach\n<!-- Describe your approach to solving the problem. -->\nThe code first checks if the **length of the input array is 3**, and if so, **returns 0.** This is because, in this case, *all elements in the array can be made equal by replacing any two elements with the third element, and hence, the sum of differences will be 0.*\nThen, the code sorts the input array using the Arrays.sort() method, which sorts the elements of the array in ascending order.\n\nAfter sorting, the code calculates three possible values for the minimized sum of absolute differences, by removing any two elements and replacing them with any element from the remaining elements in the array:\n\n- A : **removing the first two elements of the sorted array** and replacing them with any element in the middle of the array, and then calculating the absolute difference between the maximum element and the minimum element in the resulting array.\n- B : **removing the last two elements of the sorted array** and replacing them with any element in the middle of the array, and then calculating the absolute difference between the maximum element and the minimum element in the resulting array.\n- C : **removing the first and last elements of the sorted array** and replacing them with any element in the middle of the array, and then calculating the absolute difference between the maximum element and the minimum element in the resulting array.\n\nFinally, the code returns the minimum value of a, b, and c using the Math.min() method. This value represents the minimized sum of absolute differences between all pairs of elements in the input array that can be achieved by removing any two elements and replacing them with any element from the remaining elements in the array.\n\n# Complexity\n- ***Time complexity:***\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of this code is ***O(nlogn)***, where n is the length of the input array nums.\n\nThis is because the code first *sorts the input array* using Arrays.sort(nums), which has a time complexity of O(nlogn) for an array of length n.\n\nThen, the code calculates three possible values for the minimized sum of absolute differences by performing some simple arithmetic operations, which take constant time.\n\nFinally, the code returns the minimum value of the three possible values using the Math.min() method, which also takes constant time.\n\nTherefore, ***the overall time complexity of this code is dominated by the time complexity of sorting the input array, which is O(nlogn).***\n\n- ***Space complexity:***\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n**The space complexity of this code is O(1)**, which means that it uses a constant amount of extra space regardless of the size of the input array nums.\n\nThe only extra space used in this code are the integer **variables a, b, and c,** which store the three possible values of the minimized sum of absolute differences. These variables take a **constant amount of space** and do not depend on the size of the input array.\n\nIn addition, the Arrays.sort(nums) method sorts the input array in-place, without creating a new array, so it does not increase the space complexity of the code.\n\n***Therefore, the space complexity of this code is O(1).***\n\n# Code\n```\nclass Solution {\n    public int minimizeSum(int[] nums) {\n        \n        //because you can make any two elements equal to the third element and hence the array will have all elements equal and so the diffrence will be 0 for all pairs\n        \n        if(nums.length==3){\n            return 0;\n        }\n        \n        Arrays.sort(nums);\n        \n        //remove first two i.e replacing first two elements with any element in between the array\n        int a=nums[nums.length-1]-nums[2];\n        \n        //remove last two i.e replacing last two elements with any element in between the array\n        int b=nums[nums.length-3]-nums[0];\n        \n        //remove last and first i.e replacing first and last elements with any element in between the array\n        int c=nums[nums.length-2]-nums[1];\n        \n        return Math.min(a,Math.min(b,c));\n        \n        \n    }\n}\n```
['Java']