C++ || Easy Solution
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: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int M=1e9+7;\n    int dp[100001][2][3][2];\n    int f(int ind, int n, int l, int e, int t) {\n        if(ind==n) return l>=1 && e>=2 && t>=1;\n        if(dp[ind][l][e][t] != -1) return dp[ind][l][e][t];\n        long long ans=0;\n        for(char ch=\'a\';ch<=\'z\';ch++) {\n            if(ch==\'l\' and l<1) {\n                ans = ans + f(ind+1, n, l+1, e, t);\n                ans %= M;\n            }\n            else if(ch==\'e\' and e<2) {\n                ans = ans + f(ind+1, n, l, e+1, t);\n                ans %= M;\n            }\n            else if(ch==\'t\' and t<1) {\n                ans = ans + f(ind+1, n, l, e, t+1);\n                ans %= M;\n            }\n            else{\n                ans = ans + f(ind+1, n, l, e, t);\n                ans %= M;\n            }\n        }\n        return dp[ind][l][e][t] = int(ans);\n    }\n    int stringCount(int n) {\n        memset(dp, -1, sizeof(dp));\n        return f(0, n, 0, 0, 0);\n    }\n};\n```
['C++']