Python DP solution
Code\nExplanation at the bottom.\n\nB = 10**9 + 7\n\nclass Solution:\n    def stringCount(self, n: int) -> int:\n        leet = 0\n        lee = let = eet = 0\n
# Code\nExplanation at the bottom.\n```\nB = 10**9 + 7\n\nclass Solution:\n    def stringCount(self, n: int) -> int:\n        leet = 0\n        lee = let = eet = 0\n        le = ee = lt = et = 0\n        l = e = t = 0\n        everything = 1\n        for _ in range(n):\n            leet = (lee + let + eet + 23*leet) % B\n            lee = (le + ee + 24*lee) % B\n            let = (le + lt + et + 23*let) % B\n            eet = (ee + et + 24*eet) % B\n            le = (l + e + 24*le) % B\n            ee = (e + 25*ee) % B\n            lt = (l + t + 24*lt) % B\n            et = (e + t + 24*et) % B\n            l = e = t = (everything + 25*l) % B\n            everything = (everything*26) % B\n            \n        return leet    \n```\n# Approach\nAt the ith step, `leet` represents the number of strings of length `i` that have the characters to form the string `\'leet\'`. The variable `lee` similarly represents the number of strings of length `i` that have the characters to form `\'lee\'`. (Later we\'ll just say that a string "contains" `\'lee\'` for brevity.) The variables `let`, `eet`, `le`, `ee`, `lt`, `et`, `l`, `e`, and `t` are defined similarly. The variable `everything` just represents the total number of strings of length `i`, which is `26**i`.\n\nAt the ith step, we find the new value of each variable using the values of variables corresponding to shorter strings from the (i-1)th step. The general idea is to think about how many i-length strings containing a substring can end in a character `c`, for each character `c`, and then add them all up. For instance, take `leet`:\n```\nleet = <i-length strings ending with \'t\' that contain \'leet\'>\n    + <i-length strings ending with \'e\' that contain \'leet\'>\n    + <i-length strings ending with \'l\' that contain \'leet\'>\n    + <i-length strings ending with any other character that contain \'leet\'>\n```\nNote that the number of i-length strings ending with `\'t\'` that contain `\'leet\'` is equal to `lee` from the previous step. For ending character `\'e\'` this is `let` from the previous step, and for ending character `\'l\'` this is `eet` from the previous step. For any other character `c`, the number is `leet` from the previous step. There are `23` such other characters, so we multiply by `23`. Therefore we update `leet` with\n```\nleet = lee + let + eet + 23*leet\n```\nThe update rules are similar for the other variables, and we take the modulus `10**9 + 7` of everything throughout.\n\nIt\'s possible without too much extra difficulty to make this work for a general string rather than just `\'leet\'`, but solving in a way that hardcoded for this specific string was a good way to work through my thoughts during the competition, and I think it runs pretty fast and is arguably clearer this way.\n\n# Improvements\nThe above approach uses O(n) time and O(1) space, but I know of two ways to solve this problem in O(log(n)) time.\n\nThe first is with matrix exponentiation. Notice that the update we apply at each iteration of the loop is a linear recurrence. So, we can find the matrix representing this linear recurrence, raise it to the power of `n` (which can be done in O(log(n)) time), and multiply the result by the vector with a 1 in the entry corresponding to the variable `everything` and 0 for all other entries. This is very similar to how we might find fibonacci numbers in logarithmic time.\n\nThe second came from [this very good post](https://leetcode.com/problems/number-of-strings-which-can-be-rearranged-to-contain-substring/solutions/4276710/c-java-python-inclusion-exclusion-math-o-logn/). The idea is to count the number of strings which **do not** contain the characters to form `\'leet\'`, then subtract that from the total number of strings. It involves some basic combinatorics and an application of the inclusion-exclusion principle.
['Python3']