answer:Here is the corrected Python code:```pythonfrom sqlalchemy import create_engine, Table, MetaDatafrom sqlalchemy.orm import sessionmakerfrom sqlalchemy.exc import SQLAlchemyErrorimport psycopg2# ConfigurationDATABASE_URI = 'postgresql+psycopg2://user:password@localhost:5432/mydatabase'class Database: def __init__(self): self.engine = create_engine(DATABASE_URI) self.connection = self.engine.connect() self.meta = MetaData() Session = sessionmaker(bind=self.engine) self.session = Session() def create(self, table_name, kwargs): table = Table(table_name, self.meta, autoload_with=self.engine) try: insert_statement = table.insert().values(kwargs) self.connection.execute(insert_statement) except SQLAlchemyError as e: print(str(e)) self.session.rollback() else: self.session.commit() def read(self, table_name, kwargs): table = Table(table_name, self.meta, autoload_with=self.engine) select_statement = table.select().where(kwargs) result_set = self.connection.execute(select_statement) for row in result_set: print(row) def update(self, table_name, filters, kwargs): table = Table(table_name, self.meta, autoload_with=self.engine) try: update_statement = table.update().where(filters).values(kwargs) self.connection.execute(update_statement) except SQLAlchemyError as e: print(str(e)) self.session.rollback() else: self.session.commit() def delete(self, table_name, kwargs): table = Table(table_name, self.meta, autoload_with=self.engine) try: delete_statement = table.delete().where(kwargs) self.connection.execute(delete_statement) except SQLAlchemyError as e: print(str(e)) self.session.rollback() else: self.session.commit() def close(self): self.session.close() self.connection.close()```Note: I have replaced "fitens" with "filters" in the update method. I hope that was the original intent.The DATABASE_URI, user, password and other parameters such as table_name, filters and kwargs need to be actual values for this to work.

answer:Here is a simple PHP script that can display the value of a cookie for a specific key:```php<?php // Provided that the key is 'myCookieKey' key = 'myCookieKey'; if(isset(_COOKIE[key])) { echo 'Value of cookie with key ' . key . ' is: ' . _COOKIE[key]; } else { echo 'No cookie is set with key: ' . key; }?>```What this script does is check if a cookie with the key 'myCookieKey' has been set; if it has, it will print out the value of that cookie, but if it has not, it will print out a message that no cookie has been set with that key.Please make sure that the cookie is set before running this script by using `setcookie()` function in PHP.

answer:Here is a Python solution of the problem:```pythonfrom collections import Counterdef compute_disparity_and_frequency(arr1, arr2): min1, max1 = min(arr1), max(arr1) min2, max2 = min(arr2), max(arr2) disparity = max(max1-min2, max2-min1) freq = Counter(arr1+arr2) most_freq_num = freq.most_common() most_freq_num.sort(key = lambda x: (-x[1], x[0])) return disparity, most_freq_num[0][0]print(compute_disparity_and_frequency([20, 10, 4, 0, 10], [17, 1, 10, 7, 8, 7, 7]))```This will output `(20, 7)`, which corresponds to the maximum disparity and the most frequent number in both arrays.The time complexity of this solution is O(n log n), where n is the length of the array. The min, max, and Counter operations over the list are O(n), but the most time-consuming operation is the sort operation over the most common elements, which is O(n log n). The space complexity is also O(n) due to the usage of Counter. Note: The time complexity can be improved to O(n) by avoiding sort operation and finding the smallest most frequent number using linear scan, but it will increase the complexity of the code.

answer:SVG files use code to render an image, which is made up of various elements for shapes, lines, and color fills. Transforming this SVG file into a hologram projection would be quite complicated and beyond the scope of simple text-based transformation. However, if you're looking for a sci-fi vibe, you could experiment with changing the colors to something more neon or vibrant, like bright blues, pinks, or greens that often associate with sci-fi. Also, you can use gradients to create that holographic effect.You could also experiment with some CSS or JavaScript animation effects to make the image appear to move or shimmer, capturing that holographic quality. Please consider hiring a professional graphic designer or animator for more complex modifications. They would likely use a program such as Adobe Illustrator or After Effects to create a holographic effect, and could then export the file in a format suitable for your needs.

answer:SVG files can be warped or distorted using different online tools, graphic software or through directly altering the SVG file code. Here's how you do it in various ways:For Simple Online Tools:- You can use online SVG editors like vectr.com, method.ac, or vecteezy.com. Import your SVG, then use the transform/distort/warp tools to achieve your desired effect.For Graphic Software:- Adobe Illustrator: Open your SVG file with Adobe Illustrator. Select the object or path to warp using the Direct Selection Tool (A). Go to "Object > Envelope Distort > Make with Warp..." and choose your distortion settings. - Inkscape: Open your SVG file with Inkscape. Select the object with the Edit paths by nodes tool (F2). Then, use the tool options at the top: move, rotate, scale and skew to distort the object.For SVG Code Manipulation:Adjusting SVG code to warp or distort the image is more complex and requires knowledge in SVG path commands. Here's a simple example: say you've a line with the path command "M 10 10 L 20 20". To distort it, you might adjust it to "M 10 10 Q 15 0, 20 20", which adds a curve to the line.In your case, the SVG is complex with many path commands. Distorting it by changing the SVG file may be challenging without extensive knowledge of SVG paths. Using graphic software or an online tool is likely the easiest and best solution. Alternatively, you can recreate the shape and warp it during the creation process to achieve the desired effect.

answer:This problem is asking us to count the total number of sequences of length `n` that contain fewer than 2 `'A'`s and no subsequence of length 3 containing only `'L'`s. Since the number of sequences is very large and the problem asks us to return the result modulo `1e9 + 7`, we need to use dynamic programming to solve the problem in a reasonable amount of time.A good first step in tackling this problem is to categorize all possible sequences into 6 distinct states that we will use to guide our dynamic programming solution. These states basically characterize the sequences by the attendance at the last two days and the total number of absences. Here are the states we can identify:- `P...P`: ends with `P`, with no `A`- `A...P`: contains `1 A`, ends with `P`- `A...A`: contains `2 A`- `LP...P`: ends with `LP`, with no `A`- `AP...P`: contains `1 A`, ends with `LP`- `LLP...P`: ends with `LLP`, with no `A`For `n = 1` (first day), we can initialize the counts for these states. At this point, there is only one possible sequence for each of the states `-P`, `-A` and `LL-` and zero possible sequences for `-LP`, `LA-` and `LLA-`.From `n = 2` onwards, we can define a transition function to update the counts for these states based on the previous day's counts. If we denote the count after `n` days for a state `X` as `DP[X][n]`, then we can define the following transitions:- `DP[P][n] = DP[P][n-1] + DP[LP][n-1] + DP[LLP][n-1]`- `DP[A][n] = DP[P][n-1] + DP[LP][n-1] + DP[LLP][n-1]`- `DP[AP][n] = DP[A][n-1] + DP[LA][n-1] + DP[LLA][n-1]`- `DP[LP][n] = DP[P][n-1]`- `DP[LA][n] = DP[A][n-1]`- `DP[LLP][n] = DP[LP][n-1]`The total count of eligible sequences of length `n` will be the sum of all `DP[X][n]`. To get the final result, we will perform the modulo operation at each step to avoid overflow. This solution works because it enumerically evaluates all possible sequences without actually generating them, making it very efficient.Here is a Python solution implementing this approach:```pythondef checkRecord(n): mod = 109 + 7 dp = [[0, 0, 0] for _ in range(n+1)] dp[0] = [1, 1, 0] dp[1] = [2, 2, 1] for i in range(2, n+1): dp[i][0] = sum(dp[i-1]) % mod dp[i][1] = (dp[i][0]+dp[i-1][1]) % mod dp[i][2] = (dp[i][1]+dp[i-1][2]) % mod return (sum(dp[-1])*2-dp[-1][0]) % mod```Please note that this solution exploits the fact that the attendance of the day before the last day does not affect the eligibility because if the last two days are not consective lateness, then three consective lateness will not form after adding any attendance on the last day. Also, if the last two days are consective lateness, then one can only put `'P'` or `'A'` on the last day to prevent three consective lateness. Otherwise, it is not eligible. Therefore, only keep the information about the attendances on the last two days ONLY, and calculate the accumulative total amount for the three kinds of states respectively: end with `'P'`, end with `'LP'`, and end with `'LLP'` where there is NO absence day, or ONE absence day within them.