✅✅ Explanation with Proofs 💯 || C++ || ✅Clean Code ||
Intuition:\nThe problem seems to involve finding the maximum possible size of a set that can be formed using elements from two given arrays (nums1 and nums2). T
# Intuition:\nThe problem seems to involve finding the maximum possible size of a set that can be formed using elements from two given arrays (`nums1` and `nums2`). The intuition might involve using sets to represent unique elements in each array and finding the common elements between them.\n\n# Approach:\n1. Create two sets, `s1` and `s2`, to store unique elements from `nums1` and `nums2`, respectively.\n2. Create a third set, `common`, to store common elements between `s1` and `s2`.\n3. Iterate through `nums1` and insert each element into `s1`.\n4. Iterate through `nums2`, insert each element into `s2`, and check if it is common with `s1`. If common, insert it into the `common` set.\n5. Calculate the sizes of sets `s1`, `s2`, and `common` (denoted as `n1`, `n2`, and `c`).\n6. Return the minimum of the following three values:\n   - The total size of `nums1` (`n`).\n   - The minimum of (`n1 - c`) and (`n / 2`).\n   - The minimum of (`n2 - c`) and (`n / 2`).\n   - The size of the `common` set (`c`).\n# Complexity:\n- **Time Complexity:** The time complexity is O(n), where n is the total number of elements in both `nums1` and `nums2`. The code iterates through both arrays once to build the sets.\n  \n- **Space Complexity:** The space complexity is O(n), where n is the total number of elements in both `nums1` and `nums2`. The sets `s1`, `s2`, and `common` can grow linearly with the input size.\n\n# Code:\n```cpp\nclass Solution\n{\npublic:\n    int maximumSetSize(vector<int> &nums1, vector<int> &nums2)\n    {\n        int n = nums1.size();\n        unordered_set<int> s1, s2, common;\n\n        // Insert elements from nums1 into s1\n        for (auto &&i : nums1)\n        {\n            s1.insert(i);\n        }\n\n        // Insert elements from nums2 into s2 and check for common elements\n        for (auto &&i : nums2)\n        {\n            s2.insert(i);\n            if (s1.find(i) != s1.end())\n            {\n                common.insert(i);\n            }\n        }\n\n        int n1 = s1.size(), n2 = s2.size(), c = common.size();\n\n        // Return the minimum of three values\n        return min(n, min(n1 - c, n / 2) + min(n2 - c, n / 2) + c);\n    }\n};\n```
['Array', 'Hash Table', 'Greedy', 'C++']