x such that nums1 can become equal to nums2 by removing two elements and adding x to each element of nums1.

Solution

Method 1 – Sorting and Two-Pointers

Intuition

If we sort both arrays, the only difference between nums1 and nums2 is that two elements are missing from nums1 (before adding x). By trying all possible pairs of elements to remove from nums1, and checking what value of x would make the remaining elements match nums2, we can find the minimum possible x.

Approach

Sort both nums1 and nums2.
For every pair of indices (i, j) in nums1 (where i < j), remove those two elements.
For the remaining elements, calculate the difference between each element and the corresponding element in nums2.
If all differences are the same, that value is a candidate for x.
Track the minimum such x found.
Return the minimum x.

Code

[{"language":"C++","code":"class Solution {\npublic:\n    int minimumAddedInteger(vector<int>& nums1, vector<int>& nums2) {\n        sort(nums1.begin(), nums1.end());\n        sort(nums2.begin(), nums2.end());\n        int n = nums1.size(), m = nums2.size();\n        int ans = INT_MAX;\n        for (int i = 0; i < n; ++i) {\n            for (int j = i+1; j < n; ++j) {\n                vector<int> v;\n                for (int k = 0; k < n; ++k) {\n                    if (k != i && k != j) v.push_back(nums1[k]);\n                }\n                int x = nums2[0] - v[0];\n                bool ok = true;\n                for (int k = 0; k < m; ++k) {\n                    if (nums2[k] - v[k] != x) {\n                        ok = false;\n                        break;\n                    }\n                }\n                if (ok) ans = min(ans, x);\n            }\n        }\n        return ans;\n    }\n};","lang":"cpp"},{"language":"Go","code":"func minimumAddedInteger(nums1 []int, nums2 []int) int {\n    sort.Ints(nums1)\n    sort.Ints(nums2)\n    n, m := len(nums1), len(nums2)\n    ans := math.MaxInt32\n    for i := 0; i < n; i++ {\n        for j := i+1; j < n; j++ {\n            v := []int{}\n            for k := 0; k < n; k++ {\n                if k != i && k != j {\n                    v = append(v, nums1[k])\n                }\n            }\n            x := nums2[0] - v[0]\n            ok := true\n            for k := 0; k < m; k++ {\n                if nums2[k] - v[k] != x {\n                    ok = false\n                    break\n                }\n            }\n            if ok && x < ans {\n                ans = x\n            }\n        }\n    }\n    return ans\n}","lang":"go"},{"language":"Java","code":"class Solution {\n    public int minimumAddedInteger(int[] nums1, int[] nums2) {\n        Arrays.sort(nums1);\n        Arrays.sort(nums2);\n        int n = nums1.length, m = nums2.length, ans = Integer.MAX_VALUE;\n        for (int i = 0; i < n; ++i) {\n            for (int j = i+1; j < n; ++j) {\n                List<Integer> v = new ArrayList<>();\n                for (int k = 0; k < n; ++k) {\n                    if (k != i && k != j) v.add(nums1[k]);\n                }\n                int x = nums2[0] - v.get(0);\n                boolean ok = true;\n                for (int k = 0; k < m; ++k) {\n                    if (nums2[k] - v.get(k) != x) {\n                        ok = false;\n                        break;\n                    }\n                }\n                if (ok) ans = Math.min(ans, x);\n            }\n        }\n        return ans;\n    }\n}","lang":"java"},{"language":"Kotlin","cod

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