Ultra Instinct
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```cpp []\nclass Solution {\npublic:\n\n    vector<long long> dfs(int node, int prev, vector<vector<long long>>& adj, vector<long long>& ans, vector<int>& cost) {\n        vector<long long> temp2;\n\n        // Perform DFS to gather costs\n        for (auto N : adj[node]) {\n            if (N != prev) {\n                auto temp = dfs(N, node, adj, ans, cost);  // Use current node as prev\n                temp2.insert(temp2.end(), temp.begin(), temp.end());\n            }\n        }\n\n        // Include the cost of the current node\n        temp2.push_back(cost[node]);\n        sort(temp2.begin(), temp2.end());\n\n        // Handle case where not enough nodes are present\n        if (temp2.size() < 3) {\n            ans[node] = 1;\n            return temp2;  // Return the available costs\n        }\n\n        int n = temp2.size();\n        // Calculate the products using the correct elements\n        long long prd1 = temp2[n - 1] * temp2[n - 2] * temp2[n - 3]; // 3 largest\n        long long prd2 = temp2[n - 1] * temp2[0] * temp2[1];       // largest and 2 smallest\n        long long PROD = max(0LL, max(prd1, prd2)); // Ensure we use 0LL for long long\n        ans[node] = PROD;\n\n        // Return the appropriate values\n        if (temp2.size() <= 5) {\n            return temp2; // Return all available costs\n        } else {\n            // Return top 5 elements\n            return {temp2[n - 1], temp2[n - 2], temp2[n - 3], temp2[0], temp2[1]};\n        }\n    }\n\n    vector<long long> placedCoins(vector<vector<int>>& edges, vector<int>& cost) {\n        int n = cost.size();\n\n        vector<vector<long long>> adj(n);\n        for(auto& j: edges){\n            adj[j[0]].push_back(j[1]);\n            adj[j[1]].push_back(j[0]);\n        }\n\n        vector<long long> ans(n);\n        dfs(0,-1, adj, ans , cost);\n        return ans;\n    }\n};\n```
['C++']