🏆⚡️ StringView | Beat 99.25% | [C++/Java/Python] | Efficient Anagram Grouping! 🧩✨
Intuition:\nAnagrams are words or phrases formed by rearranging the letters of another word or phrase. The task is to group these anagrams together based on the
# Intuition:\nAnagrams are words or phrases formed by rearranging the letters of another word or phrase. The task is to group these anagrams together based on their sorted forms.\n\n# Approach:\n**C++ (Using string_view):**\n\n1.Initialize an unordered map with string_view as the key and a vector of strings as the value.\n2. Iterate through each string in the input array.\n3. Sort the characters of each string and store the sorted string in sorted_s.\n4. Use string_view(sorted_s) as the key to group anagrams in the map.\n5. Finally, convert the map values to a vector of vectors and return.\n\n\n**Java:**\n\n1. Initialize a HashMap with a string as the key and a list of strings as the value.\n2. Iterate through each string in the input array.\n3. Sort the characters of each string and store the sorted string in sorted_s.\n4. Use sorted_s as the key to group anagrams in the map.\n5. Finally, convert the map values to a list and return.\n\n\n**Python:**\n\n1. Initialize a defaultdict with a list as the default value type.\n2. Iterate through each string in the input list.\n3. Sort the characters of each string and join them back to form the sorted string sorted_s.\n4. Use sorted_s as the key to group anagrams in the defaultdict.\n5. Finally, convert the defaultdict values to a list and return.\n\n\n# Complexity:\n**Time Complexity: O(N * Klog(K))** where N is the number of strings and K is the length of the longest string. Sorting each string takes O(Klog(K)) time.\n**Space Complexity: O(N * K)** due to the storage of the sorted strings in the map.\n\n# Code\n\n**C++ Code:**\n\n```\nclass Solution {\npublic:\n    vector<vector<string>> groupAnagrams(vector<string>& strs) {\n        vector<vector<string>> k;\n        unordered_map<string_view, vector<string>> m; // Using string_view\n        \n        for (const string& s : strs) {\n            string sorted_s = s;\n            sort(sorted_s.begin(), sorted_s.end());\n            m[string_view(sorted_s)].push_back(s);\n        }\n        \n        for (const auto& pair : m) k.push_back(pair.second);\n        \n        return k;\n    }\n};\n```\n\n\n**Java Code:**\n\n```\nimport java.util.*;\n\nclass Solution {\n    public List<List<String>> groupAnagrams(String[] strs) {\n        Map<String, List<String>> m = new HashMap<>();\n        \n        for (String s : strs) {\n            char[] charArray = s.toCharArray();\n            Arrays.sort(charArray);\n            String sorted_s = new String(charArray);\n            if (!m.containsKey(sorted_s)) {\n                m.put(sorted_s, new ArrayList<>());\n            }\n            m.get(sorted_s).add(s);\n        }\n        return new ArrayList<>(m.values());\n    }\n}\n```\n\n\n**Python Code:**\n\n```\nclass Solution:\n    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:\n        groups = defaultdict(list)\n        for s in strs:\n            sorted_s = \'\'.join(sorted(s))\n            groups[sorted_s].append(s)\n        return list(groups.values())\n```\n\n#  Conclusion:\nAll three implementations efficiently group anagrams based on their sorted forms.\nThe C++ implementation using string_view offers an optimization by avoiding unnecessary string copies and reducing memory overhead.\nDepending on the language used and specific requirements (e.g., memory constraints, performance), one of these implementations can be chosen.
['String', 'Python', 'Java']