only prime game easy and user friendly
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n   bool solve(int n)\n    {\n        if (n <= 1)\n            return 0;\n        if (n == 2 || n == 3)\n            return 1;\n        if (n % 2 == 0 || n % 3 == 0)\n            return 0;\n        for (int i = 5; i * i <= n; i = i + 6)\n        {\n            if (n % i == 0 || n % (i + 2) == 0)\n                return 0;\n        }\n        return 1;\n    }\n\n    vector<vector<int>> findPrimePairs(int n) {\n        set<pair<int, int>> p1;\n        unordered_map<int, int> mp;\n        vector<int> v;\n        for (int i = 2; i <= n; i++) {\n            if (solve(i)) {\n                v.push_back(i);\n                mp[i]++;\n            }\n        }\n        for (int i = 0; i < v.size(); i++) {\n            if (mp.find(n - v[i]) != mp.end()) {\n                int mini = min(v[i],n-v[i]);\n                p1.insert({mini, n - mini});\n            }\n        }\n        vector<vector<int>> ans(p1.size(), vector<int>(2, 0)); // Resize the ans vector\n        int j = 0;\n        for (auto i : p1) {\n            ans[j][0] = i.first;\n            ans[j][1] = i.second;\n            j++;\n        }\n        return ans;\n    }\n};\n```
['Array', 'Math', 'C++']