✅ HashMap || Easy-Solution
\nclass Solution {\npublic:\n    bool isAnagram(string s, string t) {\n        int ns = s.length(), nt = t.length();\n        if(ns != nt) return false;\n      
```\nclass Solution {\npublic:\n    bool isAnagram(string s, string t) {\n        int ns = s.length(), nt = t.length();\n        if(ns != nt) return false;\n        unordered_map<char,int> mp;\n        for(int i = 0;i<ns;i++) mp[s[i]]++;\n        \n        for(int i = 0;i<nt;i++){\n            if(mp.find(t[i]) == mp.end() || mp[t[i]] <= 0) return false;\n            else mp[t[i]]--;\n        }\n        return true;\n    }\n};\n```
['C']