[GoLang]
dp[i][e][t][c] is the number of ways to construct a string of length i such that there are e occurences of the charactar \'e\', t occurences of the character \'
`dp[i][e][t][c]` is the number of ways to construct a string of length `i` such that there are `e` occurences of the charactar \'e\', `t` occurences of the character \'t\', and `c` occurences of the character \'c\'. Notice that if we do the DP as such, there are $O(N^4)$ states. However, we can notice that there is no difference between having two occurences of \'e\' versus three occurences of \'e\' (since we only need 2 \'e\'s), so we can just pretend that something that has >2 \'e\'s has instead 2 \'e\'s.\xA0Do the same trick for \'t\'s and \'l\'s.\n```\nfunc stringCount(n int) int {\n    dp := make([][][][]int64, n + 1)\n    for i := 0; i <= n; i++ {\n        dp[i] = make([][][]int64, 3)\n        for e := 0; e <= 2; e++ {\n            dp[i][e] = make([][]int64, 2)\n            for t := 0; t <= 1; t++ {\n                dp[i][e][t] = make([]int64, 2)\n                for c := 0; c <= 1; c++ {\n                    if i == 0 && e == 0 && t == 0 && c == 0 {\n                        dp[i][e][t][c] = 1\n                    } else if i == 0 {\n                        dp[i][e][t][c] = 0 \n                    } else {\n                        dp[i][e][t][c] += dp[i - 1][e][t][c] * 23\n                        if e != 0 {\n                            dp[i][e][t][c] += dp[i - 1][e - 1][t][c]\n                        }\n                        if e == 2 {\n                            dp[i][e][t][c] += dp[i - 1][e][t][c]\n                        }\n                        if t != 0 {\n                            dp[i][e][t][c] += dp[i - 1][e][t - 1][c]\n                        }\n                        if t == 1 {\n                            dp[i][e][t][c] += dp[i - 1][e][t][c]\n                        }\n                        if c != 0 {\n                            dp[i][e][t][c] += dp[i - 1][e][t][c - 1]\n                        }\n                        if c == 1 {\n                            dp[i][e][t][c] += dp[i - 1][e][t][c]\n                        }\n                        dp[i][e][t][c] %= int64(1e9 + 7)\n                    }\n                }\n            }\n        }\n    }\n    return int(dp[n][2][1][1])\n}\n```
['Go']