Java solution with explanation. Easy
Intuition\n Describe your first thoughts on how to solve this problem. \nIt is given that the test cases are such that there will be a x which will satisfy the 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIt is given that the test cases are such that there will be a `x` which will satisfy the given conditions.\n\nWe need to sort both the arrays in non-decreasing order and compare the elements at same index and store the difference between the elements with the window of the length of `nums2`. \n\nWe need to check that the difference of elements in the window should be equal. If few of the elements in window are not equal consider skipping the element from the `nums1 ` array. \n\nSince we are allowed to delete 2 elements from `nums1` array we can skip at most 2 elements in the window.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\nBelow is the step by step approach to solve the problem\n1. Sort the array `nums1`.\n2. Sort the array `nums2`.\n3. Calculate the number of subarrays which we can consider in the array `nums1`. We can get that by subtracting the length of `nums2` from the length of `nums1`.\n4. Initialize a variable `min` with the `Integer.MAX_VALUE`.\n5. For each sub array do the following:\n    - Initialize `k=0`, `flag=true`.\n    - Initialize `skips=2` it will be used to calculate the number of elements skipped within the subarray.\n    - Initialize `prev=null`, it will store the difference of first element of the subarray of `nums1` with the first element of `nums2`.\n    - Check the element of subarray and the array `nums2` one by one and store the difference, if the difference not equal to the prev then skip the element from the subarray and reduce the count of `skips`.\n        - We can reduce the skips value by one only if the skips is positive.\n        - Else, if we find a third element which is not equal to the  `prev` then we can discard the subarray and continue with other sub array.\n    - If the sub array was completely processed i.e. `flag=true` then check if the `min` is greater than the value of `prev`. Update the `min`.\n\n6. Return the value of `min`. \n\n# Complexity\n- Time complexity: Since, we are sorting the array. It will be `O(nlogn)`\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n- Space complexity: `O(1)`\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    public int minimumAddedInteger(int[] nums1, int[] nums2) {\n        Arrays.sort(nums1);\n        Arrays.sort(nums2);\n\n        int len1 = nums1.length - nums2.length;\n        int len2 = nums2.length;\n\n        Integer min = Integer.MAX_VALUE;\n        for (int i = 0; i <= len1; ++i) {\n            Integer prev = null;\n            boolean flag = true;\n            int k = 0;\n            int skips = 2;\n            for (int j = 0; j < len2;) {\n\n                if (i + k >= nums1.length) {\n                    flag = false;\n                    break;\n                }\n\n                if (prev == null) {\n                    prev = nums2[j] - nums1[i + k];\n                    ++j;\n                } else if (prev != (nums2[j] - nums1[i + k]) && skips <= 0) {\n                    flag = false;\n                    break;\n                } else if (prev != (nums2[j] - nums1[i + k])) {\n                    --skips;\n                } else {\n                    ++j;\n                }\n                ++k;\n            }\n            if (flag && min > prev) {\n                min = prev;\n            }\n        }\n        return min;\n    }\n}\n```
['Sliding Window', 'Sorting', 'Java']