问题

### Instructions: You are an expert Python programmer. You will be given a question (problem specification) and the first lines of Python solution to this problem, and will write in Python the remaining lines of the program to produce a correct Python program that matches the specification and passes all tests. You will NOT return anything except for the second part of the program that you wrote.
### Question:
You are given an integer n.
A string s is called good if it contains only lowercase English characters and it is possible to rearrange the characters of s such that the new string contains "leet" as a substring.
For example:

The string "lteer" is good because we can rearrange it to form "leetr" .
"letl" is not good because we cannot rearrange it to contain "leet" as a substring.

Return the total number of good strings of length n.
Since the answer may be large, return it modulo 10^9 + 7.
A substring is a contiguous sequence of characters within a string.


Example 1:

Input: n = 4
Output: 12
Explanation: The 12 strings which can be rearranged to have "leet" as a substring are: "eelt", "eetl", "elet", "elte", "etel", "etle", "leet", "lete", "ltee", "teel", "tele", and "tlee".

Example 2:

Input: n = 10
Output: 83943898
Explanation: The number of strings with length 10 which can be rearranged to have "leet" as a substring is 526083947580. Hence the answer is 526083947580 % (10^9 + 7) = 83943898.


Constraints:

1 <= n <= 10^5

### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.
```python
class Solution(object):
def stringCount(self, n):
"""
:type n: int
:rtype: int
"""
MOD = 10**9+7
return (pow(26, n, MOD)-
(25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee
```

### Answer: (enclose your partial completion in backticks. Only write the missing portion of the code, not the entire code. Be very careful to match the appropriate indentation. Directly appending your code after the partial code should produce a correct completion solution.)

答案

模型 模型答案 正确答案
金声玉亮2.0 ```
(24+24+24+n)*pow(24, n-1, MOD)+ # no l, t, e, ee
(23+23+23+n)*pow(23, n-1, MOD)+ # no l, t, e, ee
(22+22+22+n)*pow(22, n-1, MOD) # no l, t, e, ee
) % MOD
```		 class Solution(object): def stringCount(self, n): """ :type n: int :rtype: int """ MOD = 10**9+7 return (pow(26, n, MOD)- (25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee (24+24+24+n+n+0)*pow(24, n-1, MOD)- # no l|t, l|e, t|e, l|ee, t|ee, e|ee (23+n+0+0)*pow(23, n-1, MOD))%MOD # no l|t|e, l|t|ee, l|e|ee, t|e|ee
deepseek-r1

(3 * pow(24, n, MOD) + 2 * n * pow(24, n-1, MOD)) -
(pow(23, n, MOD) + n * pow(23, n-1, MOD))
) % MOD		 class Solution(object): def stringCount(self, n): """ :type n: int :rtype: int """ MOD = 10**9+7 return (pow(26, n, MOD)- (25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee (24+24+24+n+n+0)*pow(24, n-1, MOD)- # no l|t, l|e, t|e, l|ee, t|ee, e|ee (23+n+0+0)*pow(23, n-1, MOD))%MOD # no l|t|e, l|t|ee, l|e|ee, t|e|ee
gpto1 ```python
(3 * pow(24, n, MOD) + 2 * n * pow(24, n -1, MOD)) -
(pow(23, n, MOD) + n * pow(23, n -1, MOD))) % MOD
```		 class Solution(object): def stringCount(self, n): """ :type n: int :rtype: int """ MOD = 10**9+7 return (pow(26, n, MOD)- (25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee (24+24+24+n+n+0)*pow(24, n-1, MOD)- # no l|t, l|e, t|e, l|ee, t|ee, e|ee (23+n+0+0)*pow(23, n-1, MOD))%MOD # no l|t|e, l|t|ee, l|e|ee, t|e|ee
gpt-4o-2024-08-06 ```python
(n*(n-1)//2)*pow(24, n-2, MOD)) % MOD # no le, et, te
```		 class Solution(object): def stringCount(self, n): """ :type n: int :rtype: int """ MOD = 10**9+7 return (pow(26, n, MOD)- (25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee (24+24+24+n+n+0)*pow(24, n-1, MOD)- # no l|t, l|e, t|e, l|ee, t|ee, e|ee (23+n+0+0)*pow(23, n-1, MOD))%MOD # no l|t|e, l|t|ee, l|e|ee, t|e|ee
Meta-Llama-3.1-405B-Instruct-Turbo ```python
(24+24+n-1)*pow(24, n-1, MOD)+ # no l, t, e, ee, but has one of them
pow(23, n, MOD)+ # no l, t, e
(n-1)*pow(24, n-2, MOD)+ # no l, t, but has two e
(n-1)*pow(23, n-2, MOD)+ # no l, t, but has one e
(n-2)*pow(23, n-3, MOD)+ # no l, t, but has two e
(n-1)*pow(24, n-2, MOD)+ # no e, t, but has one l
(n-1)*pow(23, n-2, MOD)+ # no e, t, but has one l
(n-2)*pow(23, n-3, MOD)+ # no e, t, but has two l
(n-1)*pow(24, n-2, MOD)+ # no e, l, but has one t
(n-1)*pow(23, n-2, MOD)+ # no e, l, but has one t
(n-2)*pow(23, n-3, MOD)+ # no e, l, but has two t
(n-1)*(n-2)*pow(23, n-3, MOD)+ # no e, but has one l, one t
(n-1)*(n-2)*pow(22, n-3, MOD)+ # no e, but has one l, one t
(n-1)*(n-2)*(n-3)*pow(22, n-4, MOD) # no e, but has one l, two t
)%MOD
```		 class Solution(object): def stringCount(self, n): """ :type n: int :rtype: int """ MOD = 10**9+7 return (pow(26, n, MOD)- (25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee (24+24+24+n+n+0)*pow(24, n-1, MOD)- # no l|t, l|e, t|e, l|ee, t|ee, e|ee (23+n+0+0)*pow(23, n-1, MOD))%MOD # no l|t|e, l|t|ee, l|e|ee, t|e|ee
Meta-Llama-3.1-70B-Instruct-Turbo ```python
10*pow(24, n-2, MOD)+ # no lt, no ee
10*pow(24, n-2, MOD)+ # no le, no et
10*pow(24, n-2, MOD)+ # no ee, no lt
10*pow(24, n-2, MOD)+ # no ee, no le
10*pow(24, n-2, MOD)+ # no ee, no et
5*pow(23, n-3, MOD)+ # no lte, no e
5*pow(23, n-3, MOD)+ # no let, no e
5*pow(23, n-3, MOD)+ # no lee, no t
5*pow(23, n-3, MOD)+ # no eet, no l
pow(22, n-4, MOD)) % MOD # no leet
```		 class Solution(object): def stringCount(self, n): """ :type n: int :rtype: int """ MOD = 10**9+7 return (pow(26, n, MOD)- (25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee (24+24+24+n+n+0)*pow(24, n-1, MOD)- # no l|t, l|e, t|e, l|ee, t|ee, e|ee (23+n+0+0)*pow(23, n-1, MOD))%MOD # no l|t|e, l|t|ee, l|e|ee, t|e|ee
DeepSeek-Coder-V2-Lite-Instruct ```python
(25*25+25*2+25*2)*pow(24, n-2, MOD)+ # one l, one t, one e, one ee
(25*25+25*2)*pow(23, n-2, MOD)+ # two l, two t, two e
(25*2+25*2)*pow(22, n-2, MOD)+ # three l, three t, three e
25*pow(21, n-1, MOD)- # four l, four t
25*pow(20, n, MOD))//6 # five l, five t
```		 class Solution(object): def stringCount(self, n): """ :type n: int :rtype: int """ MOD = 10**9+7 return (pow(26, n, MOD)- (25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee (24+24+24+n+n+0)*pow(24, n-1, MOD)- # no l|t, l|e, t|e, l|ee, t|ee, e|ee (23+n+0+0)*pow(23, n-1, MOD))%MOD # no l|t|e, l|t|ee, l|e|ee, t|e|ee
Qwen2-72B-Instruct ```python
(2*24+24+24)*pow(24, n-1, MOD)- # no ll, tt, e, ee
2*23*pow(23, n-1, MOD))%MOD # no lll, tt
```		 class Solution(object): def stringCount(self, n): """ :type n: int :rtype: int """ MOD = 10**9+7 return (pow(26, n, MOD)- (25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee (24+24+24+n+n+0)*pow(24, n-1, MOD)- # no l|t, l|e, t|e, l|ee, t|ee, e|ee (23+n+0+0)*pow(23, n-1, MOD))%MOD # no l|t|e, l|t|ee, l|e|ee, t|e|ee
baichuan4 ```python
class Solution(object):
def stringCount(self, n):
"""
:type n: int
:rtype: int
"""
MOD = 10**9+7
return (pow(26, n, MOD)-
(25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee
(25+25+25+25+n)*pow(25, n-2, MOD)- # no l, t, e, ee, ee
(25+25+25+25+25+n)*pow(25, n-3, MOD) # no l, t, e, ee, ee, ee
) % MOD
```		 class Solution(object): def stringCount(self, n): """ :type n: int :rtype: int """ MOD = 10**9+7 return (pow(26, n, MOD)- (25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee (24+24+24+n+n+0)*pow(24, n-1, MOD)- # no l|t, l|e, t|e, l|ee, t|ee, e|ee (23+n+0+0)*pow(23, n-1, MOD))%MOD # no l|t|e, l|t|ee, l|e|ee, t|e|ee
doubao ```python
(24+24+24+24+n)*pow(24, n-2, MOD)- # no le, et, te
(23+23+23+n)*pow(23, n-3, MOD)+ # no lee
(22+22+n)*pow(22, n-4, MOD)- # no leet
0) % MOD
```		 class Solution(object): def stringCount(self, n): """ :type n: int :rtype: int """ MOD = 10**9+7 return (pow(26, n, MOD)- (25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee (24+24+24+n+n+0)*pow(24, n-1, MOD)- # no l|t, l|e, t|e, l|ee, t|ee, e|ee (23+n+0+0)*pow(23, n-1, MOD))%MOD # no l|t|e, l|t|ee, l|e|ee, t|e|ee
gpto1 ```python
(3 * pow(24, n, MOD) + 2 * n * pow(24, n -1, MOD)) -
(pow(23, n, MOD) + n * pow(23, n -1, MOD))) % MOD
```		 class Solution(object): def stringCount(self, n): """ :type n: int :rtype: int """ MOD = 10**9+7 return (pow(26, n, MOD)- (25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee (24+24+24+n+n+0)*pow(24, n-1, MOD)- # no l|t, l|e, t|e, l|ee, t|ee, e|ee (23+n+0+0)*pow(23, n-1, MOD))%MOD # no l|t|e, l|t|ee, l|e|ee, t|e|ee
kimi ```python
(25+25+n)*pow(25, n-2, MOD)- # no l, t, e
(25+n)*pow(25, n-3, MOD)+ # no l, t, ee
n*pow(25, n-4, MOD)) % MOD # no l, ee
```		 class Solution(object): def stringCount(self, n): """ :type n: int :rtype: int """ MOD = 10**9+7 return (pow(26, n, MOD)- (25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee (24+24+24+n+n+0)*pow(24, n-1, MOD)- # no l|t, l|e, t|e, l|ee, t|ee, e|ee (23+n+0+0)*pow(23, n-1, MOD))%MOD # no l|t|e, l|t|ee, l|e|ee, t|e|ee
minimax ```python
class Solution(object):
def stringCount(self, n):
"""
:type n: int
:rtype: int
"""
MOD = 10**9+7
return (pow(26, n, MOD) -
(25+25+25+n)*pow(25, n-1, MOD) + # no l, t, e, ee
(25+25)*pow(24, n-2, MOD) - # no l, t, e, ee, e
25*pow(23, n-3, MOD)) % MOD # no l, t, e, ee, e, e
```		 class Solution(object): def stringCount(self, n): """ :type n: int :rtype: int """ MOD = 10**9+7 return (pow(26, n, MOD)- (25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee (24+24+24+n+n+0)*pow(24, n-1, MOD)- # no l|t, l|e, t|e, l|ee, t|ee, e|ee (23+n+0+0)*pow(23, n-1, MOD))%MOD # no l|t|e, l|t|ee, l|e|ee, t|e|ee
wenxinyiyan ```python
(combinations(n-1, 2)*pow(25, n-3, MOD)) % MOD # one l, one t, two e
```

### Complete Solution with Explanation:

The complete solution, including the provided starter code and the missing portion, is as follows:

```python
class Solution(object):
def stringCount(self, n):
"""
:type n: int
:rtype: int
"""
MOD = 10**9+7
from math import pow, combinations
return (pow(26, n, MOD) -
(25+25+25+n)*pow(25, n-1, MOD) + # no l, t, e, ee
(combinations(n-1, 2)*pow(25, n-3, MOD)) % MOD # one l, one t, two e
) % MOD
```

Explanation:

1. `pow(26, n, MOD)`: This calculates the total number of possible strings of length `n` using all 26 lowercase English letters, modulo `MOD`.

2. `(25+25+25+n)*pow(25, n-1, MOD)`: This subtracts the number of strings that are missing either an 'l', 'e', 't', or have only one 'e'. There are three cases where a single letter is missing ('l', 'e', or 't'), each with `pow(25, n-1, MOD)` possibilities (since we're excluding one letter, leaving 25 to choose from). For the case of only one 'e', there are `n` positions where the single 'e' can be placed, hence the `n*pow(25, n-1, MOD)` term.

3. `combinations(n-1, 2)*pow(25, n-3, MOD)`: This adds back the number of strings that contain one 'l', one 't', and two 'e's, since these were subtracted twice in the previous step (once for missing an 'e' and once for having only one 'e'). `combinations(n-1, 2)` calculates the number of ways to choose two positions for the 'e's out of `n-1` positions (leaving one position for either 'l' or 't', which can be swapped, so we don't need to consider their order). `pow(25, n-3, MOD)` calculates the number of ways to fill the remaining `n-3` positions with any of the 25 remaining letters (excluding 'l', 't', and the two 'e's already accounted for).

4. The final result is taken modulo `MOD` to ensure it fits within the required range.

Note: The `combinations` function is imported from the `math` module to calculate the binomial coefficient "n-1 choose 2", which is the number of ways to choose 2 items from n-1 items without repetition and without order.		 class Solution(object): def stringCount(self, n): """ :type n: int :rtype: int """ MOD = 10**9+7 return (pow(26, n, MOD)- (25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee (24+24+24+n+n+0)*pow(24, n-1, MOD)- # no l|t, l|e, t|e, l|ee, t|ee, e|ee (23+n+0+0)*pow(23, n-1, MOD))%MOD # no l|t|e, l|t|ee, l|e|ee, t|e|ee
zhipu ```
(24+24+24+23+n*3)*pow(24, n-2, MOD)- # no l, t, e
(23+n*2)*pow(23, n-3, MOD)- # no l, t
(22+n)*pow(22, n-4, MOD)- # no l
n*pow(21, n-4, MOD)) % MOD # no t
```		 class Solution(object): def stringCount(self, n): """ :type n: int :rtype: int """ MOD = 10**9+7 return (pow(26, n, MOD)- (25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee (24+24+24+n+n+0)*pow(24, n-1, MOD)- # no l|t, l|e, t|e, l|ee, t|ee, e|ee (23+n+0+0)*pow(23, n-1, MOD))%MOD # no l|t|e, l|t|ee, l|e|ee, t|e|ee
deepseek-v3 ```python
(24+24+24+24+(n-1)*24)*pow(24, n-2, MOD) # no l and t, no l and e, no t and e, no e and e, no l and ee
+ (23+23+23+23)*pow(23, n-3, MOD) # no l, t, and e; no l, t, and ee; no l, e, and ee; no t, e, and ee
- 22*pow(22, n-4, MOD)) % MOD # no l, t, e, and ee
```		 class Solution(object): def stringCount(self, n): """ :type n: int :rtype: int """ MOD = 10**9+7 return (pow(26, n, MOD)- (25+25+25+n)*pow(25, n-1, MOD)+ # no l, t, e, ee (24+24+24+n+n+0)*pow(24, n-1, MOD)- # no l|t, l|e, t|e, l|ee, t|ee, e|ee (23+n+0+0)*pow(23, n-1, MOD))%MOD # no l|t|e, l|t|ee, l|e|ee, t|e|ee