Java very easy solution - Brute Force
Intuition\n1. We\'ll sort both the arrays.\n2. Take two pointers i & j as we need to remove any two elements. \n3. Leaving the elements at i & j pointers, we wi
# Intuition\n1. We\'ll sort both the arrays.\n2. Take two pointers i & j as we need to remove any two elements. \n3. Leaving the elements at i & j pointers, we will create a new array (new array will be equal to the length of nums2)\n4. We\'ll perform the subtraction for all the possible combinations of i & j pointers in nums1. \n5. Take the min possible difference.  \n\n### Time complexity: $$O(N^2M)$$\n### Space complexity: $$O(M)$$\n\n# Code\n```java []\nclass Solution {\n    public int minimumAddedInteger(int[] nums1, int[] nums2) {\n        Arrays.sort(nums1);\n        Arrays.sort(nums2);\n        int i = 0;\n        int j = 1;\n        int ans = Integer.MAX_VALUE;\n        while(i < nums1.length){\n            int [] newArray = new int [nums1.length - 2];\n            int p1 = 0;\n            for(int k = 0; k < nums1.length; k++){\n                if(p1 == newArray.length){\n                    break;\n                }\n                if(k == i || k == j){\n                    continue;\n                }\n                newArray[p1] = nums1[k];\n                p1++;\n            }\n            int check = nums2[0] - newArray[0];\n            int p2 = 0;\n            while(p2 < nums2.length){\n                int sub = nums2[p2] - newArray[p2];\n                if(sub != check){\n                    break;\n                }\n                p2++;\n            }\n            if(p2 == nums2.length){\n                ans = Math.min(ans, check);\n            }\n            if(j == nums1.length){\n                i++;\n                j = i;\n            } \n            j++;\n        }\n        return ans;\n    }\n}\n```
['Array', 'Two Pointers', 'Sorting', 'Java']