C++ Simple Hash Based Solution With Step-Wise Comments
IntuitionTry removing elems having freq>1 and than target common elements.Common elems remove from both and than later distribute to map which can accomodate it
# Intuition
Try removing elems having freq>1 and than target common elements.
Common elems remove from both and than later distribute to map which can accomodate it (i.e. removal has become negative).


# Complexity
- Time complexity: $$O(n)$$ 
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity: $$O(n)$$ 
<!-- Add your space complexity here, e.g. $$O(n)$$ -->

# Code
```cpp []
class Solution {
public:
    // Function to collect frequency of elements in a map
    void collectFrequency(const vector<int>& nums, unordered_map<int, int>& freqMap) {
        for (int num : nums) {
            freqMap[num]++;
        }
    }

    // Function to trim repeated elements until a limit 'n' is reached
    int trimRepeatedElements(unordered_map<int, int>& freqMap, int n) {
        for (auto& [key, count] : freqMap) {
            if (count > 1) {
                if (n <= 0) break;
                
                int reduction = min(n, count - 1);
                freqMap[key] -= reduction;
                n -= reduction;
            }
        }
        return n;
    }

    // Function to handle common elements between both sets
    void handleCommonElements(unordered_map<int, int>& freqMap1, unordered_map<int, int>& freqMap2, int& n1, int& n2) {
        vector<int> commonElements;
        
        // Identify common elements
        for (const auto& [key, _] : freqMap1) {
            if (freqMap2.count(key)) {
                commonElements.push_back(key);
                n1 -= freqMap1[key];
                n2 -= freqMap2[key];
            }
        }
        
        // Remove common elements from both maps
        for (int key : commonElements) {
            freqMap1.erase(key);
            freqMap2.erase(key);
        }
        
        int idx = 0;
        // Adjust remaining n1 and n2 by reintroducing common elements if necessary
        while (n1 < 0 && idx < commonElements.size()) {
            freqMap1[commonElements[idx]]++;
            idx++;
            n1++;
        }
        while (n2 < 0 && idx < commonElements.size()) {
            freqMap2[commonElements[idx]]++;
            idx++;
            n2++;
        }
    }

    // Function to trim extra elements from a frequency map
    void trimExtraElements(unordered_map<int, int>& freqMap, int& n) {
        while (n > 0 && !freqMap.empty()) {
            auto it = freqMap.begin();
            freqMap.erase(it);
            n--;
        }
    }

    // Main function to determine the maximum set size
    int maximumSetSize(vector<int>& nums1, vector<int>& nums2) {
        unordered_map<int, int> freqMap1, freqMap2, finalFreqMap;
        int n1 = nums1.size() / 2;
        int n2 = n1;

        // Step 1: Collect frequencies of elements
        collectFrequency(nums1, freqMap1);
        collectFrequency(nums2, freqMap2);

        // Step 2: Trim repeated elements
        n1 = trimRepeatedElements(freqMap1, n1);
        n2 = trimRepeatedElements(freqMap2, n2);

        // Step 3: Handle common elements
        if (n1 > 0 || n2 > 0) {
            handleCommonElements(freqMap1, freqMap2, n1, n2);
        }

        // Step 4: Trim extra elements if necessary
        if (n1 > 0) trimExtraElements(freqMap1, n1);
        if (n2 > 0) trimExtraElements(freqMap2, n2);

        // Step 5: Merge both frequency maps into final map
        for (const auto& [key, count] : freqMap1) {
            if (count > 0) finalFreqMap[key] += count;
        }
        for (const auto& [key, count] : freqMap2) {
            if (count > 0) finalFreqMap[key] += count;
        }

        return finalFreqMap.size();
    }
};

```
['C++']