explain with image
Intuition\n Describe your first thoughts on how to solve this problem. \n\n\n# Approach\n Describe your approach to solving the problem. \nbasically, res = tota
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n![Untitled-2023-09-25-1658.png](https://assets.leetcode.com/users/images/cb9275d3-746c-47f2-a320-4783cb2aa9b2_1699776058.3061638.png)\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nbasically, res = `total number of strings` - `number of strings without \'t\'` - `number of strings without \'l\'` - `number of strings without \'e\'` - `number of strings with only 1 \'e\'`\ntotal number of strings = 26^n;\nstrings without letter \'t\' = 25^n;\n...and so on...\npay attention to strings with only 1 \'e\';\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int stringCount(int n) {\n        if (n<=3) return 0;\n        if (n==4) return 12;\n        long long res=12, mod=1e9+7, total=1, noT=1,E=1, noTL=1,EnoT=1,noTLE=1, EnoTL=1;\n        for (int i=1;i<=n;i++){\n            total=total*26%mod;\n            noT=noT*25%mod;\n            noTL=noTL*24%mod;\n            noTLE=noTLE*23%mod;\n            if (i>1){\n                E= E * 25 % mod;\n                EnoT= EnoT * 24 % mod;\n                EnoTL= EnoTL * 23 % mod;\n            }\n\n        }\n        E= E * n % mod;\n        EnoT= EnoT * n % mod;\n        EnoTL= EnoTL * n % mod;\n        res=(total-noT*3-E + noTL*3 + EnoT*2-noTLE-EnoTL)%mod;\n        res=(res+mod)%mod;\n        return int(res);\n    }\n};\n```
['C++']