C++ Solution || Simple Iterative Approach Explained ✅
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n1. Sorting Arrays: The 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. **Sorting Arrays**: The first step is to sort both arrays *nums1* and *nums2* which will help in efficiently comparing elements and finding the minimum difference between elements.\n2. **Iterating through *nums1*:** We iterate through each element of *nums1* using a loop. Since the size of *nums2* is fixed, we need to consider all possible combinations of removing elements from *nums1* to match the size of *nums2*. We do this by iterating from index 0 to n - m, where n is the size of *nums1* and m is the size of *nums2*. This ensures that we have enough elements in *nums1* to compare with *nums2*.\n3. **Calculating Difference:** For each iteration, we calculate the difference (cons) between the first element of *nums2* and the current element of *nums1*. This difference represents the integer that needs to be added to elements of *nums1* to match corresponding elements of *nums2*.\n4. **Checking Validity:** We then check if all corresponding elements of *nums2* can be obtained from *nums1* after adding the calculated difference. To do this, we iterate through each element of *nums2* and check if the adjusted value (*nums2*[j] - cons) exists in *nums1*. If any element is not found, it means *nums1* doesn\'t contain enough elements to match *nums2*, so we set valid to false and break out of the loop.\n5. **Updating Minimum Difference:** If all corresponding elements can be adjusted, we update the mini variable with the minimum difference (cons) found so far. We use min() function to keep track of the smallest difference.\n6. **Returning Minimum Difference:** Finally, we return the minimum difference found. If no valid difference is found (mini remains INT_MAX), it means *nums1* already contains all elements of *nums2*, so we return 0.\n\n# Complexity\n- Time complexity: **O(N * M)**\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: **O(N)**\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int minimumAddedInteger(vector<int>& nums1, vector<int>& nums2) {\n        sort(nums1.begin(), nums1.end());\n        sort(nums2.begin(), nums2.end());\n        \n        int mini = INT_MAX;\n        int n = nums1.size();\n        int m = nums2.size();\n        \n        // Iterate through all possible combinations of removing elements\n        for (int i = 0; i <= n - m; ++i) {\n            int cons = nums2[0] - nums1[i];\n            bool valid = true;\n            vector<int> nums1_copy(nums1); // Make a copy to preserve original nums1\n            \n            // Check if all corresponding elements can be adjusted\n            for (int j = 0; j < m; ++j) {\n                auto it = find(nums1_copy.begin(), nums1_copy.end(), nums2[j] - cons);\n                if (it == nums1_copy.end()) {\n                    valid = false;\n                    break;\n                }\n                nums1_copy.erase(it); // Remove used element\n            }\n            \n            if (valid) {\n                mini = min(cons, mini);\n            }\n        }\n        \n        return mini == INT_MAX ? 0 : mini;\n    }\n};\n```
['C++']