🌟 [Explained]🔍 From Brute Force to Tries: 🚀 Mastering Sum Of Prefix Scores Optimization 🌟
\n# Approach 1: Brute Force (Memory Limit Exceeded - MLE) \uD83D\uDED1\nThis approach uses a brute-force method by storing each word\'s prefixes in a map with v
\n# Approach 1: Brute Force (Memory Limit Exceeded - MLE) \uD83D\uDED1\nThis approach uses a brute-force method by storing each word\'s prefixes in a map with vector<string>. For every prefix, the corresponding vector stores all occurrences of the prefix, leading to an excessive memory footprint.\n\n# Time Complexity:\nO(L\xB2), where L is the total length of all words combined. The repeated string copying for each prefix leads to quadratic complexity.\n# Space Complexity:\nO(L\xB2) due to storing multiple copies of each prefix.\n\n# Code\n```cpp []\nclass Solution {\npublic:\n    vector<int> sumPrefixScores(vector<string>& words) {\n        unordered_map<string, vector<string>> m;\n        for (auto const& word : words) {\n            string prefix = "";\n            for (auto const ch : word) {\n                prefix += ch;\n                m[prefix].push_back(prefix);\n            }\n        }\n        vector<int> result;\n\n        for (auto word : words) {\n            string prefix = "";\n            int count = 0;\n            for (auto ch : word) {\n                prefix += ch;\n                if (m.count(prefix) > 0) {\n                    count += m[prefix].size();\n                }\n            }\n            result.push_back(count);\n        }\n        return result;\n    }\n};\n\n```\n\n# Approach: Optimized Hash Map (Time Limit Exceeded - TLE) \u23F3\n\nThis approach improves on memory by using an unordered_map<string, int> to store prefix frequencies. However, it still suffers from performance issues due to the overhead of string operations, which can result in slow execution times for larger inputs.\n\n# Code\n```cpp []\nclass Solution {\npublic:\n    vector<int> sumPrefixScores(vector<string>& words) {\n        unordered_map<string, int> m;\n        for (auto const& word : words) {\n            string prefix = "";\n            for (auto const ch : word) {\n                prefix += ch;\n                m[prefix]++;\n            }\n        }\n        vector<int> result;\n        for (auto const& word : words) {\n            string prefix = "";\n            int count = 0;\n            for (auto const& ch : word) {\n                prefix += ch;\n                if (m.count(prefix) > 0) {\n                    count += m[prefix];\n                }\n            }\n            result.push_back(count);\n        }\n        return result;\n    }\n};\n\n\n```\n\n# Approach: [Trie-Based (Accepted) \u2705]\nThis solution uses a Trie data structure to efficiently store and query prefixes. Each node in the Trie represents a character and keeps track of how many times the prefix ending at that node has been encountered, avoiding the repeated creation of strings and optimizing both time and space usage.\n\n# Time Complexity:\nO(L), where L is the total length of all words. Each character in each word is processed only once during Trie insertion and score calculation.\n# Space Complexity:\nO(L) for storing the Trie nodes, as each character in the words is stored only once.\n\n# Code\n```cpp []\nstruct TrieNode {\n    TrieNode* children[26] = {};\n    int score = 0;\n};\nclass Trie {\npublic:\n    TrieNode* root;\n\n    Trie() { root = new TrieNode(); }\n\n    void insert(const string& word) {\n        TrieNode* node = root;\n        for (const char& ch : word) {\n            int idx = ch - \'a\';\n            if (!node->children[idx]) {\n                node->children[idx] = new TrieNode();\n            }\n            node = node->children[idx];\n            node->score++;\n        }\n    }\n\n    int getScore(const string& word) {\n        TrieNode* node = root;\n        int totalScore = 0;\n        for (const char& ch : word) {\n            int idx = ch - \'a\';\n            node = node->children[idx];\n            totalScore += node->score;\n        }\n        return totalScore;\n    }\n};\n\nclass Solution {\npublic:\n    vector<int> sumPrefixScores(vector<string>& words) {\n        Trie trie;\n        for (const auto& word : words) {\n            trie.insert(word);\n        }\n        vector<int> result;\n        for (const auto& word : words) {\n            result.push_back(trie.getScore(word));\n        }\n        return result;\n    }\n};\n\n```\n\n# Conclusion:\n# Approach 1 (MLE): \nUsing a brute-force method, storing excessive data for each prefix, leading to memory limit exceeded.\n# Approach 2 (TLE): \nAn optimized approach using a hash map, but still slow due to string operations.\n# Approach 3 (Accepted): \nThe Trie-based solution optimizes both time and space complexity, making it the most efficient approach.\n\n# \uD83C\uDF1F Enjoyed the Solutions? Found It Helpful? \uD83C\uDF1F\n# \uD83D\uDCA1 Upvote to Support the Journey from Brute Force to Optimal Solutions! \uD83D\uDCA1\n# \uD83D\uDC4D Your Upvote Helps Others Discover Efficient Approaches! \uD83D\uDE80
['Hash Table', 'String', 'Trie', 'C++']