python okay solution
Intuition\nThe problem requires finding the minimum integer \( x \) that can be added to every element of nums1 after removing exactly two elements so that it m
# Intuition\nThe problem requires finding the minimum integer \\( x \\) that can be added to every element of `nums1` after removing exactly two elements so that it matches `nums2`. We need to explore the effect of removing each pair of elements and check if adding \\( x \\) to the remaining elements results in `nums2`.\n\n# Approach\n1. **Sort the Lists**:\n   - Sort `nums1` and `nums2` to make comparison easier.\n\n2. **Brute-force with Pruning**:\n   - Iterate through all pairs of elements in `nums1` to remove.\n   - For each pair, create a temporary list of `nums1` with the pair removed.\n   - Calculate the potential \\( x \\) by subtracting the first element of `nums2` from the first element of the modified `nums1`.\n   - Check if this \\( x \\) makes the modified `nums1` equal to `nums2` when added to each element.\n   - Track the minimum \\( x \\) that satisfies the condition.\n\n3. **Return Result**:\n   - If a valid \\( x \\) is found, return it; otherwise, return -1.\n\n# Complexity\n- **Time complexity**: $$O(n^3)$$, where \\( n \\) is the length of `nums1`. This comes from iterating over all pairs of elements to remove (which is \\(O(n^2)\\)) and checking the resulting list against `nums2` (which is \\(O(n)\\)).\n- **Space complexity**: $$O(n)$$, for the temporary list and sorting.\n\n# Code\n```\nfrom typing import List\n\nclass Solution:\n    def minimumAddedInteger(self, nums1: List[int], nums2: List[int]) -> int:\n        nums2.sort()\n        min_x = float(\'inf\')\n        n = len(nums1)\n        m = len(nums2)\n        \n        if m > n - 2:\n            return -1  # If nums2 is larger than nums1 minus two elements, return -1 directly.\n        \n        nums1.sort()\n        \n        # Iterate over all possible pairs of elements to remove from nums1\n        for i in range(n):\n            for j in range(i + 1, n):\n                temp_nums1 = nums1[:i] + nums1[i+1:j] + nums1[j+1:]\n                \n                x = nums2[0] - temp_nums1[0]\n                \n                match = True\n                for k in range(m):\n                    if nums2[k] != temp_nums1[k] + x:\n                        match = False\n                        break\n                \n                if match:\n                    min_x = min(min_x, x)\n        \n        return min_x if min_x != float(\'inf\') else -1\n\n```
['Python3']