easy and simple solution.
Intuition\nbruteforce + dp + memoization\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nwe need only one \'l\' and two \'e\' and o
# Intuition\nbruteforce + dp + memoization\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nwe need only one \'l\' and two \'e\' and one \'t\' \njust use simple recursion and add memoization to it \n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n- o(n*26*1*1*2)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n- o(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nint dp[100001][3][3][3];\nint mod = 1e9+7;\nint solve(int n,int l,int e,int t)\n{\n    // cout<<n<<endl;\n    if(n==0)\n    {\n        if((l>=1)&&(e>=2)&&(t>=1))return 1;\n        return 0;\n    }\n    if(dp[n][l][e][t]!=-1)return dp[n][l][e][t];\n    int ans =0;\n    for(int i=0;i<26;i++)\n    {\n       if(i==4&&(e<2))\n       {\n           ans = (ans%mod + solve(n-1,l,e+1,t)%mod)%mod;\n       }\n       else if(i==11&&l<1)\n       {\n           ans = (ans%mod + solve(n-1,l+1,e,t)%mod)%mod;\n       }\n       else if(i==19&&t<1)\n       {\n           ans = (ans%mod + solve(n-1,l,e,t+1)%mod)%mod;\n       }\n       else\n       {\n          ans = (ans%mod + solve(n-1,l,e,t)%mod)%mod;\n       }\n    }\n    return dp[n][l][e][t] = ans;\n}\nclass Solution {\npublic:\n    int stringCount(int n) {\n        \n        memset(dp,-1,sizeof(dp));\n        return solve(n,0,0,0);\n    }\n};\n```
['C++']