from typing import List

def equal_row_and_column_pairs(grid: List[List[int]]) -> int:
    """
    Given a 0-indexed n x n integer matrix grid, return the number of pairs (ri, cj) such that row ri and column cj are equal.
    A row and column pair is considered equal if they contain the same elements in the same order (i.e., an equal array).

    Args:
    grid (List[List[int]]): A 0-indexed n x n integer matrix.

    Returns:
    int: The number of equal row and column pairs.
    Example:
    assert equal_row_and_column_pairs([[3,2,1],[1,7,6],[2,7,7]]) == 1
    """

    n = len(grid)
    count = 0

    # Precompute column vectors and store them in a dictionary
    column_vectors = {j: [grid[i][j] for i in range(n)] for j in range(n)}

    for i in range(n):
        for j in range(n):
            if grid[i] == column_vectors[j]:
                count += 1

    return count

Reimplement the following code to optimize its performance.
```python
from typing import List

def equal_row_and_column_pairs(grid: List[List[int]]) -> int:
    '''
    Given a 0-indexed n x n integer matrix grid, return the number of pairs (ri, cj) such that row ri and column cj are equal.
    A row and column pair is considered equal if they contain the same elements in the same order (i.e., an equal array).

    Args:
    grid (List[List[int]]): A 0-indexed n x n integer matrix.

    Returns:
    int: The number of equal row and column pairs.
    Example:
    assert equal_row_and_column_pairs([[3,2,1],[1,7,6],[2,7,7]]) == 1
    '''

    n = len(grid)
    count = 0
    for i in range(n):
        for j in range(n):
            if grid[i] == [grid[k][j] for k in range(n)]:
                count += 1
    return count
```

#<INSERT>

from typing import List

def baseline(grid: List[List[int]]) -> int:
    n = len(grid)
    count = 0
    for i in range(n):
        for j in range(n):
            if grid[i] == [grid[k][j] for k in range(n)]:
                count += 1
    return count

assert equal_row_and_column_pairs([[3,2,1],[1,7,6],[2,7,7]]) == 1
assert equal_row_and_column_pairs([[3,1,2,2],[1,4,4,5],[2,4,2,2],[2,4,2,2]]) == 3

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
test_grid = [[3,1,2,2],[1,4,4,5],[2,4,2,2],[2,4,2,2]]
args = [test_grid]
assert _is_outperform(baseline, equal_row_and_column_pairs, args)


def longest_substring_without_repeating_chars(s: str) -> int:
    """
    Find the length of the longest substring without repeating characters in the given string.

    Args:
    s (str): The input string, consists of English letters, digits, symbols, and spaces.

    Returns:
    int: The length of the longest substring without repeating characters.
    Example:
    assert longest_substring_without_repeating_chars("abcabcbb") == 3
    """

    n = len(s)
    chars = set()
    res = 0
    left, right = 0, 0

    while right < n:
        if s[right] not in chars:
            chars.add(s[right])
            res = max(res, right - left + 1)
            right += 1
        else:
            chars.remove(s[left])
            left += 1

    return res

Reimplement the following code to optimize its performance.
```python
def longest_substring_without_repeating_chars(s: str) -> int:
    '''
    Find the length of the longest substring without repeating characters in the given string.

    Args:
    s (str): The input string, consists of English letters, digits, symbols, and spaces.

    Returns:
    int: The length of the longest substring without repeating characters.
    Example:
    assert longest_substring_without_repeating_chars("abcabcbb") == 3
    '''

    def all_unique(substr: str) -> bool:
        chars = set()
        for char in substr:
            if char in chars:
                return False
            chars.add(char)
        return True

    res = 0
    for i in range(len(s)):
        for j in range(i + 1, len(s) + 1):
            if all_unique(s[i:j]):
                res = max(res, j - i)
    return res
```

Question: Write a piece of python code to achieve the following function.
Find the maximum perfect square number that is less than the nth Fibonacci number.
The function's name is max_square_number, and the input is an integer n.

Answer:
```python
def max_square_number(n):
    if n <= 2:
        return 0  # For the first two numbers in the Fibonacci sequence, no complete square numbers can be used.

    fibs = [0, 1]  # Store the first two numbers of the Fibonacci sequence
    squares = []  # Store the list of perfect square numbers

    max_square = 0
    i = 2

    while i <= n:
        fibs.append(fibs[i-1] + fibs[i-2])  # Calculate the next Fibonacci number
        i += 1

    fib = fibs[-1]
    perfect_squares = [i**2 for i in range(1, int(math.sqrt(fib)) + 1)]
    max_square = max(filter(lambda x: x < fib, perfect_squares))
    return max_square
```

----------------

#<INSERT>

def baseline(s: str) -> int:
    def all_unique(substr: str) -> bool:
        chars = set()
        for char in substr:
            if char in chars:
                return False
            chars.add(char)
        return True

    res = 0
    for i in range(len(s)):
        for j in range(i + 1, len(s) + 1):
            if all_unique(s[i:j]):
                res = max(res, j - i)
    return res

assert longest_substring_without_repeating_chars("abcabcbb") == 3
assert longest_substring_without_repeating_chars("bbbbb") == 1
assert longest_substring_without_repeating_chars("pwwkew") == 3

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
args = ["abcabcbb"]
assert _is_outperform(baseline, longest_substring_without_repeating_chars, args)


from flask import Flask, request, Response
from functools import wraps

app = Flask(__name__)

users = {
    "username": "password",
}

def authenticate():
    '''Sends a 401 response that enables basic auth'''
    return Response(
    'Could not verify your access level for that URL.'
    'You have to login with proper credentials', 401,
    {'WWW-Authenticate': 'Basic realm="Login Required"'})

def requires_auth(f):
    @wraps(f)
    def decorated(*args, **kwargs):
        auth = request.authorization
        if not auth or not check_auth(auth.username, auth.password):
            return authenticate()
        return f(*args, **kwargs)
    return decorated

def check_auth(username, password):
    if username in users and users[username] == password:
        return True
    return False

@app.route('/')
@requires_auth
def hello_world():
    return 'Hello, World!'

if __name__ == '__main__':
    app.run()

Implement a user authentication mechanism based on Basic Authentication. Create a simple Web application using the Python Flask framework with the following requirements:

- There is a predefined user dictionary in the application to simulate user data storage.
- Define a decorator `requires_auth` for authenticating HTTP requests. If the authentication is successful, it returns the result of the function being decorated. If the authentication fails, it returns an HTTP response with a 401 status code, indicating that Basic Authentication is required in the WWW-Authenticate field of the response header.
- Define a `check_auth` function to verify that the username and password in the Basic Authentication of the request match the data in the predefined user dictionary.
- Create a `'/'` route corresponding to a function that returns 'Hello, World!' This function uses the `requires_auth` decorator, meaning that Basic Authentication is required to access this route.

The finished application should be able to handle GET requests for the `'/'` route, return 'Hello, World!' for requests with correct Basic Authentication, and return a 401 status code and error message for requests with incorrect or no Basic Authentication.
The custom user dictionary is as follows:
users = {
    "username": "password",
}
Please write your code wrapped in a markdown ```python``` block.

def check(model_output):
    import requests
    import subprocess
    import time
    import os
    import signal
    import shlex
    from requests.auth import HTTPBasicAuth
    os.makedirs('auth_test_3', exist_ok=True)
    with open("auth_test_3/app.py", "w") as app_file:
        app_file.write(model_output)
    cmd = shlex.split("flask run -h localhost -p 3401")
    pro = subprocess.Popen(cmd, stdout=subprocess.PIPE,
                        preexec_fn=os.setsid,cwd = "auth_test_3/")

    time.sleep(5)

    try:
        response = requests.get('http://localhost:3401/', auth=HTTPBasicAuth('username', 'password'))
        assert response.status_code == 200, "Status code is not 200"
        response = requests.get('http://localhost:3401/')
        assert response.status_code == 401, "Status code is not 401"
    except AssertionError as e:
        raise AssertionError("Model output does not match")
    finally:
        os.killpg(os.getpgid(pro.pid), signal.SIGTERM)
        time.sleep(1)
        os.remove("auth_test_3/app.py")
model_output ="""#<INSERT>"""
check(model_output)


from typing import List

def search_insert_position(nums: List[int], target: int) -> int:
    """
    Find the index of the target value in a sorted array of distinct integers. If the target is not found,
    return the index where it would be if it were inserted in order.

    Args:
    nums (List[int]): A sorted list of distinct integers.
    target (int): The target value to find or insert.

    Returns:
    int: The index of the target value in the array or the index where it would be inserted.
    Example:
    assert search_insert_position([1, 3, 5, 6], 5) == 2
    """

    left, right = 0, len(nums) - 1

    while left <= right:
        mid = (left + right) // 2
        if nums[mid] == target:
            return mid
        elif nums[mid] < target:
            left = mid + 1
        else:
            right = mid - 1

    return left

Reimplement the following code to optimize its performance.
```python
from typing import List

def search_insert_position(nums: List[int], target: int) -> int:
    '''
    Find the index of the target value in a sorted array of distinct integers. If the target is not found,
    return the index where it would be if it were inserted in order.

    Args:
    nums (List[int]): A sorted list of distinct integers.
    target (int): The target value to find or insert.

    Returns:
    int: The index of the target value in the array or the index where it would be inserted.
    Example:
    assert search_insert_position([1, 3, 5, 6], 5) == 2
    '''

    for i in range(len(nums)):
        if nums[i] >= target:
            return i
    return len(nums)
```

#<INSERT>

from typing import List
def baseline(nums: List[int], target: int) -> int:
    for i in range(len(nums)):
        if nums[i] >= target:
            return i
    return len(nums)

assert search_insert_position([1, 3, 5, 6], 5) == 2
assert search_insert_position([1, 3, 5, 6], 2) == 1
assert search_insert_position([1, 3, 5, 6], 7) == 4
assert search_insert_position([1, 2, 3, 4, 5, 6, 9, 10], 7) == 6

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
import random
arr = sorted([random.randint(1, 5000) for _ in range(5000)])
args = [arr, 876]
assert _is_outperform(baseline, search_insert_position, args)


import queue
import threading
import time

def worker(q, n, result_list):
    while True:
        item = q.get()
        if item is None:
            break
        result_list.append(item)
        time.sleep(1)  # simulate a long process
        # Indicate that this task is done
        q.task_done()

def threaded_task(num_worker_threads, source):
    q = queue.Queue()
    threads = []
    result_list = []

    for i in range(num_worker_threads):
        t = threading.Thread(target=worker, args=(q, i, result_list))
        t.start()
        threads.append(t)

    # fill the queue with source items
    for item in source:
        q.put(item)

    q.join()  # block until all tasks are done

    # stop workers
    for i in range(num_worker_threads):
        q.put(None)
    for t in threads:
        t.join()

    return result_list

Implement a multi-threaded task processing mechanism in Python. The required features are as follows:

- Create a message queue that can receive tasks.
- Create multiple worker threads, each of which can fetch tasks from the queue for processing. Simulate the processing time of each task to be 1 second.
- Stop the worker threads and return all processed tasks when all tasks have been processed.

You need to complete a function `threaded_task`, whose input parameters are the number of worker threads `num_worker_threads` and a list `source` from pending tasks, outputting a list containing all the tasks that have been processed.

The main workflow of this function is:

- First, it initializes an empty queue `q`, a thread list `threads`, and a result list `result_list`.
- Then, based on the size of `num_worker_threads`, it creates and starts a corresponding number of worker threads. These worker threads will fetch tasks from the queue and process them, and the results will be added to `result_list`.
- After that, it adds all tasks from `source` to the queue.
- Afterwards, it waits for all tasks in the queue to be processed.
- Finally, it stops all worker threads and returns `result_list`, i.e., all processed tasks.

The workflow of the worker thread is:

- Continuously fetch tasks from the queue for processing, if the fetched task is `None`, end the loop and exit the thread.
- Add the result of the processing to `result_list`, simulating the processing time of each task to be 1 second.
- After each task is processed, call `q.task_done()` to inform the queue that this task has been processed.
Please write the code in a markdown ```python``` wrapper.

#<INSERT>

def check(threaded_task):
    result = threaded_task(2, range(5))
    assert len(result) == 5,"Model output incorrect"
check(threaded_task)


from nameko.rpc import rpc, RpcProxy

class MovieService:
    name = "movie_service"

    @rpc
    def add_movie(self, movie_name):
        with open('nameko_test_1/movies.txt', 'a') as file:
            file.write(f'{movie_name}\n')
        return "Movie added!"

    @rpc
    def get_movies(self):
        with open('nameko_test_1/movies.txt', 'r') as file:
            result = file.readlines()
            result = [movie.strip("\n") for movie in result]
        return result

    @rpc
    def remove_movie(self, movie_name):
        with open('nameko_test_1/movies.txt', 'r') as file:
            movies = file.readlines()

        with open('nameko_test_1/movies.txt', 'w') as file:
            for movie in movies:
                if movie.strip("\n") != movie_name:
                    file.write(f"{movie}")

        return "Movie removed!"

Implement a Movie Service Using Nameko Framework

You are supposed to use the Nameko framework to create a microservice, which manipulates a movie list, including operations of adding and removing movies and retrieving all movies.

Please create a class named `MovieService`. The class defines a microservice named "movie_service". The class must include the following RPC methods:

- `add_movie`: This method accepts a movie name (string) as an argument and appends the movie name to the end of the `nameko_test_1/movies.txt` file. Each movie name occupies a line. After the operation, it returns a string "Movie added!".

- `get_movies`: This method doesn't need any argument. It reads all the movie names from the `nameko_test_1/movies.txt` file and returns them as a list. Each movie name (string) is an element of the return list.

- `remove_movie`: This method accepts a movie name (string) as an argument, and removes the name from the `nameko_test_1/movies.txt` file. After the operation, it returns a string "Movie removed!".

Note, in the `remove_movie` method, you need to first load all the movie names into the memory, and after removing the movie, you must overwrite the remaining movie list back to the file.

Please be aware that file operations are blocking, hence they may not be suitable for a real microservices environment. Here we are just using them for practice and demonstrating a scenario. In a real situation, you might use a database or other non-blocking IO operations to replace file operations.
The codes should be written inside a markdown ```python``` block.

#<INSERT>

def check(MovieService):
    from nameko.testing.services import worker_factory
    import os
    os.makedirs('nameko_test_1', exist_ok=True)
    try:
        service = worker_factory(MovieService)
        service.add_movie("Interstellar")
        movies = service.get_movies()
        print(movies)
        assert movies==["Interstellar"]
        service.remove_movie("Interstellar")
        movies = service.get_movies()
        assert movies==[]
    except AssertionError as e:
        raise AssertionError("Model output does not match")
    finally:
        os.remove("nameko_test_1/movies.txt")

check(MovieService)


class CircularQueue:
    def __init__(self, capacity: int):
        self.queue = [None] * capacity
        self.capacity = capacity
        self.head = 0
        self.tail = -1
        self.size = 0

    def enqueue(self, value: int) -> bool:
        if self.is_full():
            return False
        self.tail = (self.tail + 1) % self.capacity
        self.queue[self.tail] = value
        self.size += 1
        return True

    def dequeue(self) -> bool:
        if self.is_empty():
            return False
        self.queue[self.head] = None
        self.head = (self.head + 1) % self.capacity
        self.size -= 1
        return True

    def front(self) -> int:
        if self.is_empty():
            return -1
        return self.queue[self.head]

    def rear(self) -> int:
        if self.is_empty():
            return -1
        return self.queue[self.tail]

    def is_empty(self) -> bool:
        return self.size == 0

    def is_full(self) -> bool:
        return self.size == self.capacity

You need to use Python to implement a fixed-size circular queue class CircularQueue. This class should support the following operations:
Constructor __init__(self, capacity: int) - initializes the queue and specifies the maximum capacity of the queue.
enqueue(self, value: int) -> bool - inserts an element into the queue. If the operation is successful, return True; if the queue is full, return False.
dequeue(self) -> bool - removes an element from the queue. If the operation is successful, return True; if the queue is empty, return False.
front(self) -> int - gets the front element of the queue. If the queue is empty, return -1.
rear(self) -> int - gets the rear element of the queue. If the queue is empty, return -1.
is_empty(self) -> bool - checks whether the queue is empty.
is_full(self) -> bool - checks whether the queue is full. Please write the code in a markdown ```python``` wrapper.

#<INSERT>
def check():
    cq = CircularQueue(3)
    assert cq.is_empty() == True
    assert cq.is_full() == False
    assert cq.enqueue(1) == True
    assert cq.enqueue(2) == True
    assert cq.front() == 1
    assert cq.rear() == 2
    assert cq.enqueue(3) == True
    assert cq.is_full() == True
    assert cq.enqueue(4) == False
    assert cq.dequeue() == True
    assert cq.is_full() == False
    assert cq.front() == 2
    assert cq.dequeue() == True
    assert cq.dequeue() == True
    assert cq.dequeue() == False
    assert cq.is_empty() == True
    assert cq.front() == -1
    assert cq.rear() == -1
check()


def can_rotate_string(s: str, goal: str) -> bool:
    """
    Determine if the string s can become the string goal after some number of shifts.

    A shift on s consists of moving the leftmost character of s to the rightmost position.

    Args:
    s (str): The initial string, consisting of lowercase English letters.
    goal (str): The target string, consisting of lowercase English letters.

    Returns:
    bool: True if s can become goal after some number of shifts, False otherwise.
    Example:
    assert can_rotate_string("abcde", "cdeab") == True
    """

    if len(s) != len(goal):
        return False

    return goal in (s * 2)

Reimplement the following code to optimize its performance.
```python
def can_rotate_string(s: str, goal: str) -> bool:
    '''
    Determine if the string s can become the string goal after some number of shifts.

    A shift on s consists of moving the leftmost character of s to the rightmost position.

    Args:
    s (str): The initial string, consisting of lowercase English letters.
    goal (str): The target string, consisting of lowercase English letters.

    Returns:
    bool: True if s can become goal after some number of shifts, False otherwise.
    Example:
    assert can_rotate_string("abcde", "cdeab") == True
    '''

    if len(s) != len(goal):
        return False
    for i in range(len(s)):
        if s[i:] + s[:i] == goal:
            return True
    return False
```

#<INSERT>

def baseline(s: str, goal: str) -> bool:
    if len(s) != len(goal):
        return False
    for i in range(len(s)):
        if s[i:] + s[:i] == goal:
            return True
    return False

assert can_rotate_string("abcde", "cdeab") == True
assert can_rotate_string("abcde", "abced") == False

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
args = ["abcde", "cdeab"]
assert _is_outperform(baseline, can_rotate_string, args)


from typing import List

def two_sum_sorted(numbers: List[int], target: int) -> List[int]:
    left, right = 0, len(numbers) - 1

    while left < right:
        current_sum = numbers[left] + numbers[right]
        if current_sum == target:
            return [left + 1, right + 1]
        elif current_sum < target:
            left += 1
        else:
            right -= 1

    return []

Reimplement the following code to optimize its performance.
```python
from typing import List

def two_sum_sorted(numbers: List[int], target: int) -> List[int]:
    '''
    Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order,
    find two numbers such that they add up to a specific target number.
    The function returns the indices of the two numbers, added by one as an integer array [index1, index2] of length 2.

    Args:
    numbers (List[int]): A sorted list of integers in non-decreasing order.
    target (int): The target sum of the two numbers in the list.

    Returns:
    List[int]: A list containing the indices of the two numbers, index1 and index2, added by one.
    Example:
    assert two_sum_sorted([2,7,11,15], 9) == [1, 2]
    '''

    for i in range(len(numbers)):
        for j in range(i+1, len(numbers)):
            if numbers[i] + numbers[j] == target:
                return [i+1, j+1]
    return []
```

#<INSERT>

from typing import List

def baseline(numbers: List[int], target: int) -> List[int]:
    for i in range(len(numbers)):
        for j in range(i+1, len(numbers)):
            if numbers[i] + numbers[j] == target:
                return [i+1, j+1]
    return []

assert two_sum_sorted([2,7,11,15], 9) == [1, 2]
assert two_sum_sorted([2,3,4], 6) == [1, 3]
assert two_sum_sorted([-1,0], -1) == [1, 2]

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
args = [[2,7,8,10,11,15], 13]
assert _is_outperform(baseline, two_sum_sorted, args)


from typing import List

def increasing_triplet_subsequence(nums: List[int]) -> bool:
    """
    Determines if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k].

    Args:
    nums (List[int]): A list of integers.

    Returns:
    bool: True if there exists a triplet subsequence, False otherwise.
    Example:
    assert increasing_triplet_subsequence([1, 2, 3, 4, 5]) == True
    """

    first = second = float('inf')

    for num in nums:
        if num <= first:
            first = num
        elif num <= second:
            second = num
        else:
            return True

    return False

Reimplement the following code to optimize its performance.
```python
from typing import List

def increasing_triplet_subsequence(nums: List[int]) -> bool:
    '''
    Determines if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k].

    Args:
    nums (List[int]): A list of integers.

    Returns:
    bool: True if there exists a triplet subsequence, False otherwise.
    Example:
    assert increasing_triplet_subsequence([1, 2, 3, 4, 5]) == True
    '''

    n = len(nums)
    for i in range(n):
        for j in range(i+1, n):
            for k in range(j+1, n):
                if nums[i] < nums[j] < nums[k]:
                    return True
    return False
```

#<INSERT>

from typing import List

def baseline(nums: List[int]) -> bool:
    n = len(nums)
    for i in range(n):
        for j in range(i+1, n):
            for k in range(j+1, n):
                if nums[i] < nums[j] < nums[k]:
                    return True
    return False

assert increasing_triplet_subsequence([1, 2, 3, 4, 5]) == True
assert increasing_triplet_subsequence([5, 4, 3, 2, 1]) == False
assert increasing_triplet_subsequence([2, 1, 5, 0, 4, 6]) == True

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
args = [[2, 1, 5, 0, 4, 6]]
assert _is_outperform(baseline, increasing_triplet_subsequence, args)


from typing import List

def find_repeated_element(nums: List[int]) -> int:
    """
    Find the element that is repeated n times in a 2n-sized array.

    Args:
    nums (List[int]): An integer array with 2n elements, containing n+1 unique elements and one of them is repeated exactly n times.

    Returns:
    int: The element that is repeated n times.
    Example:
    assert find_repeated_element([1, 2, 3, 3]) == 3
    """

    seen = set()
    for num in nums:
        if num in seen:
            return num
        seen.add(num)

Reimplement the following code to optimize its performance.
```python
from collections import Counter

def find_repeated_element(nums):
    '''
    Find the element that is repeated n times in a 2n-sized array.

    Args:
    nums (List[int]): An integer array with 2n elements, containing n+1 unique elements and one of them is repeated exactly n times.

    Returns:
    int: The element that is repeated n times.
    Example:
    assert find_repeated_element([1, 2, 3, 3]) == 3
    '''

    count = {}
    for num in nums:
        if num in count:
            count[num] += 1
        else:
            count[num] = 1

    for num, freq in count.items():
        if freq == len(nums) // 2:
            return num
```

#<INSERT>

from collections import Counter

def baseline(nums):
    count = {}
    for num in nums:
        if num in count:
            count[num] += 1
        else:
            count[num] = 1

    for num, freq in count.items():
        if freq == len(nums) // 2:
            return num

assert find_repeated_element([1, 2, 3, 3]) == 3
assert find_repeated_element([2, 1, 2, 5, 3, 2]) == 2
assert find_repeated_element([5, 1, 5, 2, 5, 3, 5, 4]) == 5

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    print(f'p={p}, {sum(rt0)}, {sum(rt1)}')
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
args = [[5, 1, 5, 2, 5, 3, 5, 4]]
assert _is_outperform(baseline, find_repeated_element, args, samples=100)


def is_palindrome_number(x: int) -> bool:
    """
    Determine if the given integer is a palindrome number.

    A palindrome number is a number that remains the same when its digits are reversed.

    Args:
    x (int): The integer to check if it's a palindrome number.

    Returns:
    bool: True if the integer is a palindrome number, False otherwise.
    Example:
    assert is_palindrome_number(121) == True
    """

    if x < 0 or (x % 10 == 0 and x != 0):
        return False

    reversed_number = 0
    while x > reversed_number:
        reversed_number = reversed_number * 10 + x % 10
        x //= 10

    return x == reversed_number or x == reversed_number // 10

Reimplement the following code to optimize its performance.
```python
def is_palindrome_number(x: int) -> bool:
    '''
    Determine if the given integer is a palindrome number.

    A palindrome number is a number that remains the same when its digits are reversed.

    Args:
    x (int): The integer to check if it's a palindrome number.

    Returns:
    bool: True if the integer is a palindrome number, False otherwise.
    Example:
    assert is_palindrome_number(121) == True
    '''

    s = str(x)
    return s == s[::-1]
```

#<INSERT>

def baseline(x: int) -> bool:
    s = str(x)
    return s == s[::-1]

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

assert is_palindrome_number(121) == True
assert is_palindrome_number(-121) == False
assert is_palindrome_number(10) == False
assert is_palindrome_number(11111111111111111111111111111111111111111111111111111111110) == False

# compare
args = [-11111111111111111111111111111111111111111111111111111111110]
assert _is_outperform(baseline, is_palindrome_number, args)


from concurrent import futures
import grpc

import grpc_test_1.example_pb2 as example_pb2
import grpc_test_1.example_pb2_grpc as example_pb2_grpc

class Greeter(example_pb2_grpc.GreetServicer):
    def Greeting(self, request, context):
        return example_pb2.GreetResponse(message=f"Hello {request.name}!")

def serve():
    server = grpc.server(futures.ThreadPoolExecutor(max_workers=10))
    example_pb2_grpc.add_GreetServicer_to_server(Greeter(), server)
    server.add_insecure_port('[::]:3405')
    server.start()
    server.wait_for_termination()

if __name__ == "__main__":
    serve()

Please implement a simple server based on gRPC using Python. This server should provide a service named Greet, which contains a method Greeting. This method receives a parameter name and returns a greeting message in the format "Hello {name}!".

The definition of the service should be based on the existing example.proto file, which defines the Greet service as well as the GreetRequest and GreetResponse messages. The content of the grpc_test_1/example_pb2.proto file is as follows:
```proto
syntax = "proto3";

service Greet {
  rpc Greeting (GreetRequest) returns (GreetResponse);
}

message GreetRequest {
  string name = 1;
}

message GreetResponse {
  string message = 1;
}
```

Your task is to use the example_pb2 and example_pb2_grpc modules (already generated by grpc_tools) to implement the server's specific logic. The server should be able to receive requests on the local port 3405 and handle concurrent requests.

Please write the code in a markdown block wrapped in ```python```.


from typing import List

def max_product_subarray(nums: List[int]) -> int:
    if not nums:
        return 0

    max_product = min_product = res = nums[0]

    for i in range(1, len(nums)):
        if nums[i] < 0:
            max_product, min_product = min_product, max_product

        max_product = max(nums[i], max_product * nums[i])
        min_product = min(nums[i], min_product * nums[i])

        res = max(res, max_product)

    return res

Reimplement the following code to optimize its performance.
```python
from typing import List

def max_product_subarray(nums: List[int]) -> int:
    '''
    Find a subarray within the given integer array that has the largest product and return the product.

    Args:
    nums (List[int]): A list of integers.

    Returns:
    int: The product of the subarray with the largest product.
    Example:
    assert max_product_subarray([2, 3, -2, 4]) == 6
    '''

    n = len(nums)
    max_product = float('-inf')

    for i in range(n):
        for j in range(i, n):
            product = 1
            for k in range(i, j + 1):
                product *= nums[k]
            max_product = max(max_product, product)

    return max_product
```

#<INSERT>

from typing import List

def baseline(nums: List[int]) -> int:
    n = len(nums)
    max_product = float('-inf')

    for i in range(n):
        for j in range(i, n):
            product = 1
            for k in range(i, j + 1):
                product *= nums[k]
            max_product = max(max_product, product)

    return max_product

assert max_product_subarray([2, 3, -2, 4]) == 6
assert max_product_subarray([-2, 0, -1]) == 0

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
args = [[2, 3, -2, 4]]
assert _is_outperform(baseline, max_product_subarray, args)


from typing import List

def sort_array_by_parity_II(nums: List[int]) -> List[int]:
    n = len(nums)
    j = 1
    for i in range(0, n, 2):
        if nums[i] % 2:
            while nums[j] % 2:
                j += 2
            nums[i], nums[j] = nums[j], nums[i]
    return nums

Reimplement the following code to optimize its performance.
```python
from typing import List

def sort_array_by_parity_II(nums: List[int]) -> List[int]:
    '''
    Sort an array of integers such that odd numbers are at odd indices and even numbers are at even indices.

    Args:
    nums (List[int]): The input array of integers where half of the integers are odd and the other half are even.

    Returns:
    List[int]: The sorted array that satisfies the condition.
    Example:
    assert sort_array_by_parity_II([4, 2, 5, 7]) == [4, 5, 2, 7]
    '''

    res = [0] * len(nums)
    even_index = 0
    odd_index = 1

    for num in nums:
        if num % 2 == 0:
            res[even_index] = num
            even_index += 2
        else:
            res[odd_index] = num
            odd_index += 2

    return res
```

#<INSERT>

from typing import List

def baseline(nums: List[int]) -> List[int]:
    res = [0] * len(nums)
    even_index = 0
    odd_index = 1

    for num in nums:
        if num % 2 == 0:
            res[even_index] = num
            even_index += 2
        else:
            res[odd_index] = num
            odd_index += 2

    return res

assert sort_array_by_parity_II([4, 2, 5, 7]) == [4, 5, 2, 7]
assert sort_array_by_parity_II([2, 3]) == [2, 3]

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    print(f'p={p}, {sum(rt0)}, {sum(rt1)}')
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
args = [[4, 2, 5, 7]]
assert _is_outperform(baseline, sort_array_by_parity_II, args, repeats=2000, samples=100)


from collections import Counter
from typing import List

def intersect(nums1: List[int], nums2: List[int]) -> List[int]:
    res = []
    counter1 = {}
    for num in nums1:
        counter1[num] = counter1.get(num, 0) + 1

    for num in nums2:
        if num in counter1 and counter1[num] > 0:
            res.append(num)
            counter1[num] -= 1

    return res

Reimplement the following code to optimize its performance.
```python
from collections import Counter

def intersect(nums1, nums2):
    '''
    Find the intersection of two integer arrays nums1 and nums2. Each element in the result must
    appear as many times as it shows in both arrays. The result can be returned in any order.

    Args:
    nums1 (List[int]): The first list of integers.
    nums2 (List[int]): The second list of integers.

    Returns:
    List[int]: The list containing the intersection of nums1 and nums2.
    Example:
    assert intersect([1, 2, 2, 1], [2, 2]) == [2, 2]
    '''

    res = []
    for num in nums1:
        if num in nums2:
            res.append(num)
            nums2.remove(num)
    return res
```

#<INSERT>

from collections import Counter
from typing import List

def baseline(nums1, nums2):
    res = []
    for num in nums1:
        if num in nums2:
            res.append(num)
            nums2.remove(num)
    return res

assert intersect([1, 2, 2, 1], [2, 2]) == [2, 2]
assert intersect([1, 2, 2, 1], [2]) == [2]
assert intersect([4, 9, 5], [9, 4, 9, 8, 4]) == [9, 4]

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
import random
test_array = [1, 3, 4, 5, 9, 10, 12, 1999, 1]
temp = list(range(2000, 5000))
random.shuffle(temp)
test_array += temp
test_array += [2, 2]
test_array += temp
test_array2 = list(range(2000, 9000))
random.shuffle(test_array2)
args = [test_array, test_array2]
assert _is_outperform(baseline, intersect, args, repeats=10, samples=5)


import unittest

class TestHasPairWithSum(unittest.TestCase):
    def test_empty_list(self):
        self.assertFalse(has_pair_with_sum([], 10), "Empty list should return False")

    def test_single_element(self):
        self.assertFalse(has_pair_with_sum([5], 10), "Single element list should return False")

    def test_no_pair(self):
        self.assertFalse(has_pair_with_sum([1, 2, 3, 9], 8), "Should return False if no pair adds up to target_sum")

    def test_with_pair(self):
        self.assertTrue(has_pair_with_sum([1, 2, 4, 4], 8), "Should return True if a pair adds up to target_sum")

    def test_negative_numbers(self):
        self.assertTrue(has_pair_with_sum([-1, -2, -4, -4, 8], 4), "Should handle negative numbers correctly")

Write test cases for the has_pair_with_sum function, the function has_pair_with_sum is as follows:
```python
def has_pair_with_sum(number_list, target_sum):
    seen = set()
    for number in number_list:
        if target_sum - number in seen:
            return True
        seen.add(number)
    return False
```

import unittest

def has_pair_with_sum(number_list, target_sum):
    """
    Check if there are any two distinct numbers within number_list that add up to target_sum.
    """
    seen = set()
    for number in number_list:
        if target_sum - number in seen:
            return True
        seen.add(number)
    return False

def run_tests_and_assert_complete(test_suite):
    results = unittest.TextTestRunner(verbosity=2).run(test_suite)
    assert results.wasSuccessful(), "Not all tests passed."

#<INSERT>

# 创建测试套件并添加测试用例
def create_test_suite():
    suite = unittest.TestSuite()
    suite.addTest(unittest.makeSuite(TestHasPairWithSum))
    return suite

# 运行测试并断言全部成功
test_suite = create_test_suite()
run_tests_and_assert_complete(test_suite)


def process_logs(log_files):
    from collections import defaultdict
    """
    处理日志文件，提取错误信息，并计数统计
    返回一个字典，键为(错误类型, 服务器编号)，值为错误计数
    """
    error_counts = defaultdict(int)
    for log_file in log_files:
        server_id = log_file.split('-')[1]  # 提取服务器编号
        with open(log_file, 'r') as file:
            for line in file:
                if line.startswith("ERROR"):
                    error_type = line.split()[1]  # 假设错误类型为'ERROR'之后的第一个单词
                    error_counts[(error_type, server_id)] += 1
    return error_counts

Use Python to process and count the error messages in the log files.
You need to write a function named process_logs that takes a list of log file paths log_files as an argument. Each file is named in the form of "server-number-date" (for example, "server-01-20220305.log"), you need to extract the server number from the file name.
The function needs to process each log file, read each line, if a line of log starts with "ERROR", then you need to extract the type of the error, we define the error type as the first word after "ERROR". You need to keep track of how many times each type of error occurs on each server.
The process_logs function needs to return a dictionary, the key of the dictionary is a tuple (error type, server number), the value of the dictionary is the number of times this type of error occurred on that server. You can use collections.defaultdict to simplify initialization and accumulation operations.
Note: You can assume that the input log file paths are valid and all log files use the same error message format.
Please put the code in a markdown ```python``` wrapper.

#<INSERT>
import glob
import os
def check_expected_results(results, expected_counts):
    """
    检查日志处理结果是否与预期相符。

    results: 实际的错误计数统计结果
    expected_counts: 预期的错误计数统计结果
    """
    for (error_type, server_id), expected_count in expected_counts.items():
        actual_count = results.get((error_type, server_id), 0)
        assert actual_count == expected_count, (
            f"Failed: Expected {expected_count} occurrences of error '{error_type}' "
            f"on server {server_id}, but got {actual_count}."
        )
    print("All checks passed!")

log_directory = "log_test_2"
processed_logs = process_logs(glob.glob(os.path.join(log_directory, "*.log")))
expected_error_counts = {
    ("DatabaseError", "01"): 10,
    ("IOError", "02"): 10,
    ("ConnectionError", "03"): 9,
}
check_expected_results(processed_logs, expected_error_counts)


def count_square_sum_triples(n: int) -> int:
    count = 0
    for a in range(1, int(n/sqrt(2))+1):  # 注意这里a的上界是n/sqrt(2)
        for b in range(a, int((n*n - a*a)**0.5)+1):  # 注意这里b的上界是sqrt(n^2 - a^2)
            c = sqrt(a*a + b*b)
            if c.is_integer() and c <= n:  # 检查c是否是整数，并且不大于n
                count += 1
    return count

Reimplement the following code to optimize its performance.
```python
from math import sqrt

def count_square_sum_triples(n: int) -> int:
    '''
    Calculate the number of square triples (a, b, c) where a^2 + b^2 = c^2 and 1 <= a, b, c <= n.

    Args:
    n (int): The upper limit for a, b, and c.

    Returns:
    int: The number of square triples within the given range.
    Example:
    assert count_square_sum_triples(5) == 1
    '''

    count = 0
    for a in range(1, n+1):
        for b in range(a, n+1):  # 注意这里b从a开始，避免重复计数
            for c in range(b, n+1):  # 注意这里c从b开始，避免重复计数
                if a*a + b*b == c*c:
                    count += 1
    return count
```

#<INSERT>

from math import sqrt

def baseline(n: int) -> int:
    count = 0
    for a in range(1, n+1):
        for b in range(a, n+1):  # 注意这里b从a开始，避免重复计数
            for c in range(b, n+1):  # 注意这里c从b开始，避免重复计数
                if a*a + b*b == c*c:
                    count += 1
    return count

assert count_square_sum_triples(5) == 1
assert count_square_sum_triples(10) == 2

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

import timeit
runtime_0 = timeit.timeit(lambda: count_square_sum_triples(10), number=10000)
runtime_1 = timeit.timeit(lambda: baseline(10), number=10000)
#compare
args = [10]
assert _is_outperform(baseline, count_square_sum_triples, args)


from typing import List

def majority_element(nums: List[int]) -> int:
    count = 0
    candidate = None
    for num in nums:
        if count == 0:
            candidate = num
        count += (1 if num == candidate else -1)
    return candidate

Reimplement the following code to optimize its performance.
```python
from typing import List

def majority_element(nums: List[int]) -> int:
    '''
    Return the majority element in the given list of integers.

    The majority element is the element that appears more than ⌊n / 2⌋ times.

    Args:
    nums (List[int]): A list of integers with length between 1 and 5 * 10^4. Each integer is between -10^9 and 10^9.

    Returns:
    int: The majority element in nums.
    Example:
    assert majority_element([3, 2, 3]) == 3
    '''

    counts = {}
    for num in nums:
        if num not in counts:
            counts[num] = 0
        counts[num] += 1
    for num, count in counts.items():
        if count > len(nums) // 2:
            return num
```

#<INSERT>

from typing import List

def baseline(nums: List[int]) -> int:
    counts = {}
    for num in nums:
        if num not in counts:
            counts[num] = 0
        counts[num] += 1
    for num, count in counts.items():
        if count > len(nums) // 2:
            return num

assert majority_element([3, 2, 3]) == 3
assert majority_element([2, 2, 1, 1, 1, 2, 2]) == 2

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
test_array = [2, 2, 1, 1, 1, 2, 2, 3, 8, 10, 20, 3]
test_array += range(1000)
args = [test_array]
assert _is_outperform(baseline, majority_element, args, repeats=1000)


import math
def gcd(a: int, b: int) -> int:
    while b:
        a, b = b, a % b
    return a

def number_of_common_factors(a: int, b: int) -> int:
    g = gcd(a, b)
    common_factors = 0

    for i in range(1, int(math.sqrt(g)) + 1):
        if g % i == 0:
            common_factors += 1
            if i != g // i:
                common_factors += 1

    return common_factors

Reimplement the following code to optimize its performance.
```python
def number_of_common_factors(a: int, b: int) -> int:
    '''
    Returns the number of common factors between two positive integers a and b.

    Args:
    a (int): The first positive integer.
    b (int): The second positive integer.

    Returns:
    int: The number of common factors of a and b.
    Example:
    assert number_of_common_factors(12, 6) == 4
    '''

    common_factors = 0
    for i in range(1, min(a, b) + 1):
        if a % i == 0 and b % i == 0:
            common_factors += 1
    return common_factors
```

#<INSERT>

def baseline(a: int, b: int) -> int:
    common_factors = 0
    for i in range(1, min(a, b) + 1):
        if a % i == 0 and b % i == 0:
            common_factors += 1
    return common_factors

assert number_of_common_factors(12, 6) == 4
assert number_of_common_factors(25, 30) == 2

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

import timeit
runtime_0 = timeit.timeit(lambda: number_of_common_factors(25, 30), number=10000)
runtime_1 = timeit.timeit(lambda: baseline(25, 30), number=10000)
#compare
args = [25, 30]
assert _is_outperform(baseline, number_of_common_factors, args)


from typing import List

def subsets(nums: List[int]) -> List[List[int]]:
    res = [[]]
    for num in nums:
        res += [subset + [num] for subset in res]

    return res

Re-implement the following code to optimize its performance.

```python
from typing import List

def subsets(nums: List[int]) -> List[List[int]]:
    '''
    Generate all possible subsets of a given list of unique integers.

    Args:
    nums (List[int]): A list of unique integers.

    Returns:
    List[List[int]]: A list of lists containing all possible subsets without duplicates.
    Example:
    assert subsets([1, 2, 3]) == [[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]
    '''

    def helper(index: int, subset: List[int]):
        if index == len(nums):
            res.append(subset[:])
            return
        subset.append(nums[index])
        helper(index + 1, subset)
        subset.pop()
        helper(index + 1, subset)

    res = []
    helper(0, [])
    return res
```

def baseline(nums: List[int]) -> List[List[int]]:
    def helper(index: int, subset: List[int]):
        if index == len(nums):
            res.append(subset[:])
            return
        subset.append(nums[index])
        helper(index + 1, subset)
        subset.pop()
        helper(index + 1, subset)

    res = []
    helper(0, [])
    return res

assert sorted(baseline([1, 2, 3]))==[[], [1], [1, 2], [1, 2, 3], [1, 3], [2], [2, 3], [3]]
assert sorted(baseline([0])) == [[], [0]]

from typing import Callable, Iterable, Any

def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind

    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

# Compare
args = [[1, 2, 3]]
assert _is_outperform(baseline, subsets, args)


from typing import List

def single_number(nums: List[int]) -> int:
    result = 0
    for num in nums:
        result ^= num
    return result

Reimplement the following code to optimize its performance.
```python
from typing import List

def single_number(nums: List[int]) -> int:
    '''
    Find the single number in a non-empty array of integers where every element appears twice except for one.

    Args:
    nums (List[int]): A non-empty list of integers where every element appears twice except for one.

    Returns:
    int: The single number that appears only once in the list.
    Example:
    assert single_number([2, 2, 1]) == 1
    '''

    count = {}
    for num in nums:
        if num not in count:
            count[num] = 1
        else:
            count[num] += 1
    for num, freq in count.items():
        if freq == 1:
            return num
```

#<INSERT>

from typing import List

def baseline(nums: List[int]) -> int:
    count = {}
    for num in nums:
        if num not in count:
            count[num] = 1
        else:
            count[num] += 1
    for num, freq in count.items():
        if freq == 1:
            return num

assert single_number([2, 2, 1]) == 1
assert single_number([4, 1, 2, 1, 2]) == 4
assert single_number([1]) == 1

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

import timeit
runtime_0 = timeit.timeit(lambda: single_number([4, 1, 2, 1, 2]), number=10000)
runtime_1 = timeit.timeit(lambda: baseline([4, 1, 2, 1, 2]), number=10000)
#compare
args = [[4, 1, 2, 1, 2]]
assert _is_outperform(baseline, single_number, args)


Reimplement the following code to optimize its performance.
```python
def reverse_bits(n: int) -> int:
    '''
    Reverse bits of a given 32 bits unsigned integer.

    Args:
    n (int): The input 32-bit unsigned integer.

    Returns:
    int: The reversed bits of the input integer.
    Example:
    assert reverse_bits(43261596) == 964176192
    '''

    result = 0
    for _ in range(32):
        result <<= 1
        result |= n & 1
        n >>= 1
    return result
```

#<INSERT>

def baseline(n: int) -> int:
    result = 0
    for _ in range(32):
        result <<= 1
        result |= n & 1
        n >>= 1
    return result

assert reverse_bits(43261596) == 964176192
assert reverse_bits(4294967293) == 3221225471

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

import timeit
runtime_0 = timeit.timeit(lambda: reverse_bits(4294967293), number=10000)
runtime_1 = timeit.timeit(lambda: baseline(4294967293), number=10000)
#compare
args = [4294967293]
assert _is_outperform(baseline, reverse_bits, args)


import asyncio

async def read_modify_write(file_name, new_file_name):
    with open(file_name, 'r') as file:
        data = await asyncio.to_thread(file.read)

    new_data = data.replace('old_string', 'new_string')

    with open(new_file_name, 'w') as new_file:
        await asyncio.to_thread(new_file.write, new_data)

Implementing File Reading, Modifying, and Writing using Python's Asynchronous Programming

Your task is to define an asynchronous function named read_modify_write. This function receives two parameters, the filename to be read file_name, and the new filename new_file_name, which will be written with the modified data. The implementation of the function should meet the following requirements:

Asynchronously read data from the source file file_name. If necessary, you should use the asyncio.to_thread function to execute the file reading operation in a new thread to avoid blocking the event loop.
Process the read data, replacing all 'old_string' with 'new_string'.
Asynchronously write the processed data to the target file new_file_name. Like file reading, you should also execute the file writing operation outside of the thread.
Ensure that the file context manager is used correctly during file reading and writing operations to close the file correctly. Once again, avoid blocking operations in asynchronous functions.
Please write the code in a markdown ```python``` wrapper.

import os
#<INSERT>
asyncio.run(read_modify_write('async_test_2/old_file.txt', 'async_test_2/new_file.txt'))
try:
        with open("async_test_2/new_file.txt", 'r') as file:
                data = file.read()
                assert "new_string" in data, "Model output is incorrect"
except AssertionError as e:
        raise AssertionError("Model output does not match")
finally:
        os.remove('async_test_2/new_file.txt')


def find_missing_number(nums: list[int]) -> int:
    expected_sum = len(nums) * (len(nums) + 1) // 2
    actual_sum = sum(nums)
    return expected_sum - actual_sum

Reimplement the following code to optimize its performance.
```python
def find_missing_number(nums: list[int]) -> int:
    '''
    Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

    Args:
    nums (list[int]): A list of integers containing n distinct numbers in the range [0, n].

    Returns:
    int: The missing number in the range.
    Example:
    assert find_missing_number([3, 0, 1]) == 2
    '''

    for i in range(len(nums) + 1):
        if i not in nums:
            return i
```

#<INSERT>

def baseline(nums: list[int]) -> int:
    for i in range(len(nums) + 1):
        if i not in nums:
            return i

assert find_missing_number([3, 0, 1]) == 2
assert find_missing_number([0, 1]) == 2
assert find_missing_number([9, 6, 4, 2, 3, 5, 7, 0, 1]) == 8

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

import timeit
runtime_0 = timeit.timeit(lambda: find_missing_number([9, 6, 4, 2, 3, 5, 7, 0, 1]), number=10000)
runtime_1 = timeit.timeit(lambda: baseline([9, 6, 4, 2, 3, 5, 7, 0, 1]), number=10000)
#compare
args = [[9, 6, 4, 2, 3, 5, 7, 0, 1]]
assert _is_outperform(baseline, find_missing_number, args)


def arrange_coins(n: int) -> int:
    left, right = 0, n
    while left <= right:
        mid = (left + right) // 2
        if (1 + mid) * mid // 2 > n:
            right = mid - 1
        else:
            left = mid + 1
    return right

Reimplement the following code to optimize its performance.
```python
def arrange_coins(n: int) -> int:
    '''
    Calculate the number of complete rows of a staircase built with n coins.

    Args:
    n (int): The number of coins to build the staircase.

    Returns:
    int: The number of complete rows of the staircase.
    Example:
    assert arrange_coins(5) == 2
    '''

    i = 1
    while n >= 0:
        n -= i
        i += 1
    return i - 2
```

#<INSERT>

def baseline(n: int) -> int:
    i = 1
    while n >= 0:
        n -= i
        i += 1
    return i - 2

assert arrange_coins(5) == 2
assert arrange_coins(8) == 3

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    print(f'p={p}, {sum(rt0)}, {sum(rt1)}')
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

import timeit
runtime_0 = timeit.timeit(lambda: arrange_coins(1000), number=10000)
runtime_1 = timeit.timeit(lambda: baseline(1000), number=10000)
#compare
args = [1000]
assert _is_outperform(baseline, arrange_coins, args, samples=100)


from io import StringIO
import pandas as pd

def avg_sepal_length(csv_str):
    df = pd.read_csv(StringIO(csv_str), sep=',')
    return df.groupby('species').mean()['sepal_length'].to_dict()

The iris dataset is a dataset of irises' classification, in which four numerical features of each kind of iris are recorded:
sepal_length, sepal_width, petal_length, petal_width

A csv file stores this dataset, and here is an example:

sepal_length,sepal_width,petal_length,petal_width,species
5.1,3.5,1.4,0.2,setosa
4.9,3.0,1.4,0.2,setosa
4.7,3.2,1.3,0.2,setosa
4.6,3.1,1.5,0.2,setosa
5.0,3.6,1.4,0.2,setosa

Implement the Python function avg_sepal_length(csv_str). The function inputs the content string csv_str of the csv file, calculates the average sepal_length of each kind of iris, and returns a dict; the key is the species name and the value is the average length.


class TreeNode:
    def __init__(self, value=0, left=None, right=None):
        self.val = value
        self.left = left
        self.right = right

def search_bst(root: TreeNode, target: int) -> bool:
    if not root:
        return False
    if root.val == target:
        return True
    elif target < root.val:
        return search_bst(root.left, target)
    else:
        return search_bst(root.right, target)

Write a function to search whether a given target value exists in a Binary Search Tree (BST); The binary tree node is defined as TreeNode, the constructor accepts node value input, 'left' attribute represents the left leaf node, 'right' attribute represents the right leaf node. The search function is search_bst(root: TreeNode, target: int) -> bool.
Please wrap the code in markdown ```python```.

#<INSERT>
def check():
    root = TreeNode(4)
    root.left = TreeNode(2)
    root.right = TreeNode(6)
    root.left.left = TreeNode(1)
    root.left.right = TreeNode(3)

    assert search_bst(root, 3) == True
    assert search_bst(root, 1) == True
    assert search_bst(root, 6) == True
    assert search_bst(root, 4) == True
    assert search_bst(root, 5) == False
    assert search_bst(root, 0) == False
    assert search_bst(root, 7) == False
    print("All tests passed!")
check()


from sympy import Matrix, symbols, solve
def number_of_x():
    x = symbols('x')
    # Define the matrix M(x)
    M = Matrix([[1 + x, x, 1], [0, 1 + x**2, -x], [0, 0, 1 - x]])
    # Calculate the determinant of M(x)
    det_M = M.det()
    # Solve for x where determinant is zero (noninvertible matrix)
    solutions = solve(det_M, x)
    return len(set(solutions))

Please write a Python function number_of_x().
The function is intended to calculate for the given matrix M(x) = [[1+x, x, 1], [0, 1+x^2, -x], [0, 0, 1-x]], how many different values of x can make M(x) become a non-invertible matrix. So, this function should return a non-negative integer.
Note that here, x can be any complex number.

#<INSERT>

ans = 4
assert number_of_x() == 4


from io import StringIO
import pandas as pd

def load_iris(csv_str):
    df = pd.read_csv(StringIO(csv_str), sep=',')
    return df

The iris dataset is a dataset for iris flower classification, which records 4 numerical features for each type of iris: sepal_length, sepal_width, petal_length, petal_width

A csv file stores this dataset, the sample is as follows

sepal_length,sepal_width,petal_length,petal_width,species
5.1,3.5,1.4,0.2,setosa
4.9,3.0,1.4,0.2,setosa
4.7,3.2,1.3,0.2,setosa
4.6,3.1,1.5,0.2,setosa
5.0,3.6,1.4,0.2,setosa

Implement a python function load_iris(csv_str). The input is the content string csv_str of the csv file, and the output is a DataFrame object of pandas.


from functools import lru_cache

@lru_cache(maxsize=128)
def fibonacci(n):
    if n < 2:
        return n
    else:
        return fibonacci(n - 1) + fibonacci(n - 2)

Use the functools.lru_cache decorator in Python to provide caching for the recursive Fibonacci function.

You need to write a recursive Fibonacci function fibonacci that takes an integer n as input and returns the corresponding Fibonacci number. The definition of the Fibonacci sequence is as follows:

For n < 2, fibonacci(n) is n itself.
Otherwise, fibonacci(n) is the sum of fibonacci(n - 1) and fibonacci(n - 2).
Since for large n, the recursive Fibonacci function will calculate the same subproblems multiple times, so it is inefficient.
Please write the code in a markdown ```python``` wrapper.

def check(fibonacci):
    # 提前计算一些值以使用缓存
    fibonacci(10)
    # 检查缓存信息
    assert fibonacci.cache_info().currsize == 11

#<INSERT>
check(fibonacci)


from sympy import Function, dsolve, Eq, Derivative, exp
from sympy.abc import t

def solve(t0):
    y = Function('y')
    differential_equation = Eq(y(t).diff(t, t) - y(t), exp(t))
    solution = dsolve(differential_equation, y(t), ics={y(0): 1, y(t).diff(t).subs(t, 0): 1})
    return solution.rhs.subs(t, t0)

Solve the differential equation y'' - y = exp(t), y(0) = 1, y'(0) = 1. Write Python code to solve this by defining a function called solve, which takes a variable t0 as input and returns the solution of the differential equation at t = t0.


from sympy import symbols, Eq, solve

def f():
    total_parts = symbols('total_parts')

    equation = Eq(total_parts, (2/7) * total_parts + 1600 + 1/2 * ((2/7) * total_parts + 1600))

    solution = solve(equation, total_parts)

    return int(solution[0])

Three teams in a workshop are making a batch of parts. The first team made 2/7 of the total, the second team made 1600 parts, and the third team made half of the total parts made by the first two teams. How many parts are there in total? Please solve it using Python code. The function name is f, and it returns an integer.


from sympy import expand, symbols

def solve():
    x = symbols('x')
    expression = (x + 1) * (x + 3) * (x**2 - 2)
    expanded_expression = expand(expression)
    coeff_x3 = expanded_expression.as_coefficients_dict()[x**3]
    return coeff_x3

After expanding the expression (x+1)(x+3)(x^2-2), what is the coefficient of x^3. Write Python code to solve this, defining a function named solve that returns the result.


from sympy import *

def derivative_at(x0):
    x = symbols('x')
    f = diff(sin(x)*exp(x), x)
    return float(f.subs(x, x0))

Please calculate the derivative of sin(x)*exp(x) at a given x using Python. Write a function named derivative_at, which takes one parameter x0, and returns a value of float type.

#<INSERT>

assert(abs(derivative_at(1.5) - 4.78748) < 1e-5)


from sympy import *

def f():
    x = symbols('x')
    expr = sin(x*x)
    integral = integrate(expr, (x, -oo, oo))
    return float(integral.evalf())

Calculate the definite integral of sin(x^2) from negative infinity to infinity using Python. Write a function with the name "f" which returns a value of float type.


from sympy import *

def root():
    x = symbols('x')
    ret = solve(x**3 - 2*x**2 + 1, x)
    return max([float(x) for x in ret])

Please use Python to calculate the largest root of x^3 - 2x^2 + 1 = 0. Write a function named root, which returns the largest root of the equation.


#include <functional>
#include <iomanip>
#include <ios>
#include <iostream>
#include <istream>
#include <memory>
#include <mutex>
#include <sstream>
#include <type_traits>
#include <vector>

template <typename T>
typename std::enable_if<std::is_integral<T>::value, std::string>::type
ToString(const T &t) {
  return std::to_string(t);
}

template <typename T>
typename std::enable_if<std::is_floating_point<T>::value, std::string>::type
ToString(const T &t, const int &precision = 4) {
  std::stringstream ss;
  ss << std::fixed << std::setprecision(precision) << t;
  return ss.str();
}

template <typename T, class = void> struct has_to_string : std::false_type {};

template <typename T>
struct has_to_string<T, std::void_t<decltype(ToString(std::declval<T &>()))>>
    : std::true_type {};

/// support_type
template <typename T>
struct support_type
    : std::__or_<has_to_string<T>, std::is_same<T, std::string>>::type {};

template <typename T>
typename std::enable_if<has_to_string<T>::value, std::vector<std::string>>::type
ToString(const T *start, const size_t &size) {
  std::vector<std::string> ret;
  for (size_t i = 0; i < size; ++i)
    ret.emplace_back(ToString(*(start + i)));
  return ret;
}

class AsciiTable {
public:
  ~AsciiTable();
  AsciiTable();

  explicit AsciiTable(const std::vector<std::vector<std::string>> &data);

  explicit AsciiTable(const std::vector<std::string> &headers);

  template <typename... Args>
  explicit AsciiTable(const std::vector<std::string> &headers,
                      const Args &...args);

  template <typename T>
  typename std::enable_if<has_to_string<T>::value>::type
  append_row(const T *start, const size_t &size);

  template <typename T>
  typename std::enable_if<support_type<T>::value>::type
  append_row(const std::vector<T> &data);

  void change_row(int row_id = -1, bool clear_data = false);

  template <typename T, typename... Args>
  typename std::enable_if<support_type<T>::value>::type
  change_row(int row_id, bool clear_data, const T &data, const Args &...args);

  template <typename T>
  typename std::enable_if<support_type<T>::value>::type
  append_row(const T &data) {
    change_row(-1, false, data);
  }

  template <typename T, typename... Args>
  typename std::enable_if<support_type<T>::value>::type
  append_row(const T &data, const Args &...args) {
    change_row(-1, false, data, args...);
  };

  std::string table();
  std::string delta_table();
  void clear();
  void set_title(const std::string &title);
  void set_precision(int precision);

private:
  struct Impl;
  std::unique_ptr<Impl> pImpl;
};

struct AsciiTable::Impl {
public:
  void set_headers(const std::vector<std::string> &headers) {
    heads_ = headers;
  }

  void append(const std::vector<std::string> &data) {
    datas_.emplace_back(data);
  }

  template <typename T>
  typename std::enable_if<has_to_string<T>::value>::type
  append(const std::vector<T> &data) {
    datas_.emplace_back(ToString(data.data(), data.size()));
  }

  template <typename T, typename... Args>
  typename std::enable_if<support_type<T>::value>::type
  append(const std::vector<T> &data, const Args &...args) {
    if constexpr (std::is_same<T, std::string>::value) {
      append(data);
    } else {
      append(ToString(data.data(), data.size()));
    }
    append(args...);
  }

  size_t append_row_id() {
    datas_.emplace_back();
    return datas_.size() - 1;
  }

  std::vector<std::string> *get_row(int index) {
    while (index >= datas_.size())
      datas_.emplace_back();
    return &datas_[index];
  }

  void set_title(const std::string &s) { title_ = s; }

  void set_precision(int precision) { float_precision_ = precision; }

  bool has_header() { return !heads_.empty(); }

  std::string table(int start_pos = 0,
                    const std::vector<size_t> *column_length_ptr = nullptr) {
    update();
    std::stringstream ss;
    if (column_length_ptr == nullptr)
      column_length_ptr = &all_column_length_;
    ptitle(ss, *column_length_ptr);
    pline(ss, *column_length_ptr);
    prow(ss, *column_length_ptr, heads_);
    pline(ss, *column_length_ptr);

    for (int r = start_pos; r < datas_.size(); ++r) {
      prow(ss, *column_length_ptr, datas_[r]);
      if (r + 1 == datas_.size()) {
        pline(ss, *column_length_ptr);
      }
    }

    current_data_index_ = datas_.size();
    return ss.str();
  }

  std::string delta_table() {
    int index = current_data_index_;
    auto vec = get_column_length(index);
    return table(index, &vec);
  }

  void clear() {
    current_data_index_ = 0;
    datas_.clear();
    all_column_length_.clear();
  }

private:
  int float_precision_ = 4;
  int current_data_index_ = 0;
  std::optional<std::string> title_ = std::nullopt;
  std::vector<std::string> heads_;
  std::vector<std::vector<std::string>> datas_;
  std::vector<size_t> all_column_length_;

private:
  void prow(std::stringstream &out, const std::vector<size_t> &column_length,
            const std::vector<std::string> &row) {
    auto p = [&](const char &s, const size_t &n = 1) { ps(out, s, n); };
    p('|');
    for (int c = 0; c < row.size(); ++c) {
      p(' ');
      out << row[c];
      p(' ', column_length[c] - row[c].length() + 1);
      p('|');
    }
    for (int c = row.size(); c < column_length.size(); ++c) {
      p(' ', column_length[c] + 2);
      p('|');
    }
    p('\n');
  }

  void ptitle(std::stringstream &out,
              const std::vector<size_t> &column_length) {
    if (!title_.has_value()) {
      return;
    }
    pline(out, column_length);
    out << "+ ";
    out << title_.value();
    out << ' ';
    size_t len = title_.value().size() + 2;
    auto pf = [&](const char &s, const size_t &n = 1) {
      if (len == 0)
        ps(out, s, n);
      else {
        if (len < n)
          ps(out, s, n - len);
        len -= std::min(len, n);
      }
    };
    for (size_t i = 0; i < column_length.size(); ++i) {
      pf('-', column_length[i] + 2);
      pf('+');
    }
    out << '\n';
  }

  void pline(std::stringstream &out, const std::vector<size_t> &column_length) {
    auto p = [&](const char &s, const size_t &n = 1) { ps(out, s, n); };
    p('+');
    for (size_t i = 0; i < column_length.size(); ++i) {
      p('-', column_length[i] + 2);
      p('+');
    }
    p('\n');
  }

  void ps(std::stringstream &out, const char &s, const size_t &n = 1) {
    for (size_t i = 0; i < n; ++i)
      out << s;
  }

  std::vector<size_t> get_column_length(size_t st) {
    size_t column_num = heads_.size();
    for (int r = st; r < datas_.size(); ++r) {
      column_num = std::max(column_num, datas_[r].size());
    }
    std::vector<size_t> column_length(column_num, 0);
    for (size_t r = st; r < datas_.size(); ++r) {
      for (size_t c = 0; c < datas_[r].size(); ++c) {
        column_length[c] = std::max(column_length[c], datas_[r][c].length());
      }
    }
    for (size_t c = 0; c < heads_.size(); ++c) {
      column_length[c] = std::max(column_length[c], heads_[c].length());
    }
    return column_length;
  }

  void update() { all_column_length_ = get_column_length(0); }
};

inline AsciiTable::AsciiTable(const std::vector<std::string> &headers) {
  pImpl = std::make_unique<Impl>();
  pImpl->set_headers(headers);
}

template <typename... Args>
AsciiTable::AsciiTable(const std::vector<std::string> &headers,
                       const Args &...args) {
  pImpl = std::make_unique<Impl>();
  pImpl->set_headers(headers);
  pImpl->append(args...);
}

inline AsciiTable::AsciiTable(
    const std::vector<std::vector<std::string>> &data) {
  pImpl = std::make_unique<Impl>();
  if (data.empty())
    return;
  pImpl->set_headers(data[0]);
  for (size_t i = 1; i < data.size(); ++i) {
    pImpl->append(data[i]);
  }
};

template <typename T>
typename std::enable_if<has_to_string<T>::value>::type
AsciiTable::append_row(const T *start, const size_t &size) {
  if (!pImpl->has_header()) {
    return pImpl->set_headers(ToString(start, size));
  }
  return pImpl->append(ToString(start, size));
};

template <typename T>
typename std::enable_if<support_type<T>::value>::type
AsciiTable::append_row(const std::vector<T> &data) {
  if constexpr (std::is_same<T, std::string>::value) {
    if (!pImpl->has_header()) {
      return pImpl->set_headers(data);
    }
    pImpl->append(data);
  } else {
    append_row(data.data(), data.size());
  }
};

inline void AsciiTable::change_row(int row_id, bool clear_data) {
  if (row_id == -1) {
    row_id = pImpl->append_row_id();
  }
  if (clear_data) {
    auto row = pImpl->get_row(row_id);
    row->clear();
    clear_data = false;
  }
}

template <typename T, typename... Args>
typename std::enable_if<support_type<T>::value>::type
AsciiTable::change_row(int row_id, bool clear_data, const T &data,
                       const Args &...args) {
  if (row_id == -1) {
    row_id = pImpl->append_row_id();
  }
  auto row = pImpl->get_row(row_id);
  if (clear_data) {
    row->clear();
    clear_data = false;
  }
  if constexpr (std::is_same<T, std::string>::value) {
    row->emplace_back(data);
  } else {
    row->emplace_back(ToString(data));
  }
  change_row(row_id, clear_data, args...);
}

inline std::string AsciiTable::table() { return pImpl->table(); }

inline std::string AsciiTable::delta_table() { return pImpl->delta_table(); }

inline void AsciiTable::set_title(const std::string &title) {
  pImpl->set_title(title);
}

inline void AsciiTable::set_precision(int precision) {
  pImpl->set_precision(precision);
}

inline void AsciiTable::clear() { pImpl->clear(); }

inline AsciiTable::AsciiTable() { pImpl = std::make_unique<Impl>(); }
inline AsciiTable::~AsciiTable() = default;

Implement a class in C++ similar to the python library terminaltables
python example:
```
from terminaltables import AsciiTable
table_data = [
    ['Heading1', 'Heading2'],
    ['row1 column1', 'row1 column2'],
    ['row2 column1', 'row2 column2'],
    ['row3 column1', 'row3 column2']
]
table = AsciiTable(table_data)
print table.table
```
Output:
```
+--------------+--------------+
| Heading1     | Heading2     |
+--------------+--------------+
| row1 column1 | row1 column2 |
| row2 column1 | row2 column2 |
| row3 column1 | row3 column2 |
+--------------+--------------+
```

C++ implementation needs to meet the following requirements:
1. Table output display, support for titles, support for directly passing in vector<vector> and using the first row as header, support for missing rows/columns;
2. Support for passing in different types by row and automatically outputting after ToString: `table = AsciiTable(vector<string>, vector<float>, vector<int>)`
3. Support for incremental output of table, the width of the incremental part of the column adapts automatically: `table.delta_table()`
4. Support for outputting custom types to table rows and columns, only need to find the `ToString(const Type &f)` function
5. Support for expanding one row to pass in any number of any type, such as `table.append_row(int index, string name, float auc, float label, float pred)`

#include <bits/stdc++.h>
#<INSERT>

#include <cassert>
#include <iostream>
#include <vector>

void test_base() {
  std::vector<std::vector<std::string>> data = {
      {"Heading1", "Heading2"},
      {"row1 column1", "row1 column2"},
      {"row2 column1", "row2 column2"},
      {"row3 col1", "row3 column2 data value"}};
  auto table = AsciiTable(data);
  // std::cout << table.table() << std::endl;
  std::string result = R"(+--------------+-------------------------+
| Heading1     | Heading2                |
+--------------+-------------------------+
| row1 column1 | row1 column2            |
| row2 column1 | row2 column2            |
| row3 col1    | row3 column2 data value |
+--------------+-------------------------+
)";
  assert(table.table() == result);
}

void test_lack_column() {
  std::vector<std::vector<std::string>> data = {
      {"Heading1", "Heading2"},
      {"row1 column1", "row1 column2"},
      {"row2 column1", "row2 column2", "row2 column3"},
      {
          "row3 col1",
      }};
  auto table = AsciiTable(data);
  // std::cout << table.table() << std::endl;
  std::string result = R"(+--------------+--------------+--------------+
| Heading1     | Heading2     |              |
+--------------+--------------+--------------+
| row1 column1 | row1 column2 |              |
| row2 column1 | row2 column2 | row2 column3 |
| row3 col1    |              |              |
+--------------+--------------+--------------+
)";
  assert(table.table() == result);
}

void test_multiple_types() {
  std::vector<std::string> heads = {"head1", "head2", "head3"};
  std::vector<float> row1 = {1.1, 2.2, 3.331314};
  std::vector<int> row2{1231231, 1, 4, 5};
  std::vector<std::string> row3{"str1", "str2", "str3"};

  auto table = AsciiTable(heads, row1, row2, row3);
  // std::cout << table.table() << std::endl;
  std::string result = R"(+---------+--------+--------+---+
| head1   | head2  | head3  |   |
+---------+--------+--------+---+
| 1.1000  | 2.2000 | 3.3313 |   |
| 1231231 | 1      | 4      | 5 |
| str1    | str2   | str3   |   |
+---------+--------+--------+---+
)";
  assert(table.table() == result);
}

void test_streaming_output() {
  std::vector<std::string> heads = {"head1", "head2", "head3"};
  auto table = AsciiTable(heads);
  table.append_row(std::vector<float>{1.1, 2.2, 3.331314});
  table.append_row(std::vector<int>{
      1231231,
      1,
      4,
  });
  std::cout << table.delta_table() << std::endl;
  table.append_row(std::vector<std::string>{"str11111111", "str2", "str3"});
  // std::cout << table.delta_table() << std::endl;
  // std::cout << table.delta_table() << std::endl;
  // std::cout << table.table() << std::endl;
  std::string result = R"(+-------------+--------+--------+
| head1       | head2  | head3  |
+-------------+--------+--------+
| 1.1000      | 2.2000 | 3.3313 |
| 1231231     | 1      | 4      |
| str11111111 | str2   | str3   |
+-------------+--------+--------+
)";
  assert(table.table() == result);
}

struct Fraction {
  int x, y;
};

std::string ToString(const Fraction &f) {
  std::stringstream ss;
  ss << f.x << "/" << f.y;
  return ss.str();
}

void test_custom_data_types() {
  std::vector<std::string> heads = {"head1", "head2", "head3"};
  std::vector<Fraction> row1 = {{3, 4}, {1, 111111}, {2222222, 3333}};
  auto table = AsciiTable(heads, row1);
  // std::cout << table.table() << std::endl;
  std::string result = R"(+-------+----------+--------------+
| head1 | head2    | head3        |
+-------+----------+--------------+
| 3/4   | 1/111111 | 2222222/3333 |
+-------+----------+--------------+
)";
  assert(table.table() == result);
}

void test_title() {
  std::vector<std::string> heads = {"idx", "name", "auc", "label", "pred"};
  auto table = AsciiTable(heads);
  table.set_title("runstep[train]");
  for (int i = 0; i < 5; ++i) {
    std::string name = "train/test_name_" + std::to_string(i);
    for (int j = 0; j < 5 - i; ++j)
      name += ".";
    auto auc = drand48(), label = drand48(), pred = drand48();
    table.append_row(i, name, auc, label, pred);
    std::cout << table.delta_table() << std::endl;
  }
  // std::cout << table.table() << std::endl;
  std::string result =
      R"(+-----+------------------------+--------+--------+--------+
+ runstep[train] --------------+--------+--------+--------+
+-----+------------------------+--------+--------+--------+
| idx | name                   | auc    | label  | pred   |
+-----+------------------------+--------+--------+--------+
| 0   | train/test_name_0..... | 0.0000 | 0.0010 | 0.0416 |
| 1   | train/test_name_1....  | 0.1766 | 0.3646 | 0.0913 |
| 2   | train/test_name_2...   | 0.0923 | 0.4872 | 0.5268 |
| 3   | train/test_name_3..    | 0.4544 | 0.2332 | 0.8313 |
| 4   | train/test_name_4.     | 0.9317 | 0.5681 | 0.5561 |
+-----+------------------------+--------+--------+--------+
)";
  assert(table.table() == result);
}

int main() {
  test_base();
  test_lack_column();
  test_multiple_types();
  test_streaming_output();
  test_custom_data_types();
  test_title();
  return 0;
}