Intuitive approach - Best solution
Here I have added the brute force and the Optimal solution for the problem:\n\n# Brute Force\n\n# Intuition\n Describe your first thoughts on how to solve this 
# Here I have added the brute force and the Optimal solution for the problem:\n\n# Brute Force\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFor any array, if we are deleting any element, then the rest of the elements except the deleted one will hold the mentioned relationship.\n\nFor eg: lets say we have an array ```num1 = [1 2 3 4]``` and an array ```num2 = [2 3 4 5]```.\n\nNow, let\'s say I remove ```5``` from num2. Still, if I look at the elements ```2, 3, 4```, they still hold relationship with elements ```1, 2, 3``` of num1. I.e ```2 - 1 = 1```, ```3 - 1 = 2``` and ```4 - 1 = 3```.\n\nTherefore, what we have to do is that we will sort both the arrays so that we get the numbers in a sequence. After doing this, we will caluclate the difference ```diff``` between the ```ith``` element of ```num1``` and the ```jth``` element of ```num2```. Then we will count the number of elements in    ```num1``` that on adding ```diff``` gives me the ```jth``` element of num2.\n\nIf the current element of ```num1``` is not equal to the ```jth``` element of ```num2```, then I increment the index for ```num1``` else I increment the index for both ```num1``` and ```num2```.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe code tackles the problem by iteratively comparing elements in the sorted arrays ```nums1``` and ```nums2```. Here\'s a breakdown of the steps:\n\n**Sorting:**\n\nBoth ```nums1``` and ```nums2``` are sorted using ```Arrays.sort()```. This allows for efficient element comparison.\n\n**Outer Loop:**\n\nAn outer loop `(for (int i = 0; i < nums1.length; i++))` iterates through each element in nums1.\n\n**Target Difference:**\n\nThe initial difference `(diff)` is calculated as the first element of `nums2` minus the current element `(nums1[i])` in `nums1`. This represents the potential value of x needed to make the elements equal.\n\n**Inner Loop:**\n\nAn inner loop `(while (idx < nums1.length && idx1 < nums2.length))` iterates simultaneously through `nums1` and `nums2`.\n\n**Matching Elements:**\n\nIf the current element in `nums1`, `(nums1[idx])` plus the `diff` is equal to the current element in `nums2`, `(nums2[idx1])`, it indicates that `x` is a possible value. Both indices `(idx and idx1)` are incremented, and a counter `(count)` is incremented to track matching elements.\n\n**Mismatch:**\n\nIf there\'s a mismatch, only the index in `nums1 (idx)` is incremented, as `nums2` is assumed to have the correct elements due to sorting. This ensures that nums1 doesn\'t skip elements while searching for a match.\n\n**Potential Solution:**\n\nIf the count (number of matches) becomes equal to the length of nums2 `(nums2.length)`, it means all elements in `nums2` have been found in nums1 with the potential diff (value of x).\n\n**Minimum Update:**\n\nThe ans variable `(initially Integer.MAX_VALUE)` is updated with the minimum of its current value and the `diff`. This keeps track of the smallest possible x that achieves the equivalence.\n\n**Return Minimum:**\n\nAfter iterating through all elements in nums1, the ans variable is returned. If ans remains `Integer.MAX_VALUE`, it signifies that no valid `x` was found, and `0` is returned.\n# Complexity\n- Time complexity: ***O(N*(N-2))**\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\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 count = 0;\n        int ans = Integer.MAX_VALUE;\n        for(int i = 0; i < nums1.length; i ++){\n            int idx = i;\n            int idx1 = 0;\n            int diff = nums2[idx1] - nums1[idx];\n            count = 0;\n\n            while(idx < nums1.length && idx1 < nums2.length){\n                if(nums1[idx] + diff == nums2[idx1]){\n                    idx ++;\n                    idx1 ++;\n                    count ++;\n                }\n                else idx ++;\n            }\n\n            if(count == nums2.length)\n                ans = Math.min(ans, diff);\n        }\n\n        return ans == Integer.MAX_VALUE? 0: ans;\n    }\n}\n```\n\n# Best Solution\n\n# Intuition\nHere, we will be taking the difference between the forst element of `nums2` and the first 3 elements of `nums1`. Accordingly, we calculate the validity for all three and return the minimum.\n\n# Complexity\n- Time complexity: ***O(NLogN + 3 x N) i.e O(NLogN)***\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: ***O(1)***\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n```\nclass Solution {\n    private boolean isValid(int []nums1, int []nums2, int diff){\n        int count = 0;\n        int j = 0;\n        for(int i = 0; i < nums1.length; i ++){\n            if(j == nums2.length) break;\n            if(nums2[j] - nums1[i] == diff){\n                j ++;\n                count ++;\n            }\n        }\n\n        if(count == nums2.length) return true;\n\n        return false;\n    }\n    public int minimumAddedInteger(int[] nums1, int[] nums2) {\n        Arrays.sort(nums1);\n        Arrays.sort(nums2);\n\n        int ans = Integer.MAX_VALUE;\n        int diff1 = nums2[0] - nums1[0];\n        int diff2 = nums2[0] - nums1[1];\n        int diff3 = nums2[0] - nums1[2];\n\n        if(isValid(nums1, nums2, diff1)) ans = Math.min(ans, diff1);\n\n        if(isValid(nums1, nums2, diff2)) ans = Math.min(ans, diff2);\n\n        if(isValid(nums1, nums2, diff3)) ans = Math.min(ans, diff3);\n\n        return ans == Integer.MAX_VALUE? 0: ans;\n    }\n}\n```
['Java']