Easy soln-->using frequency count with map
\n\n# Code\ncpp []\nclass Solution {\npublic:\n    int lengthAfterTransformations(string s, int t) {\n        vector<int>ans(26,0);\n       \n        map<char,i
\n\n# Code\n```cpp []\nclass Solution {\npublic:\n    int lengthAfterTransformations(string s, int t) {\n        vector<int>ans(26,0);\n       \n        map<char,int>m;\n        for(int i=0;i<s.size();i++){\n            m[s[i]]++;\n        }\n        for(auto i:m){\n            ans[i.first-\'a\']=i.second;\n        }\n        for(int i=0;i<t;i++){\n            long long hold=ans[25];\n            for(int i=25;i>=1;i--){\n                ans[i]=ans[i-1];\n            }\n            ans[0]=hold;\n            ans[1]=(ans[1]+hold)%1000000007;\n        }\n        long long count=0;\n        for(int i=0;i<26;i++){\n            count=(count+ans[i])%1000000007;\n        }\n        return count;\n\n        \n        \n    }\n};\n```
['C++']