Python OOPE Exam

Let's solve your problem step by step!

Let's create a fun Python program for your task! 😃

📝 Function: within_and_has_double_quotes Let’s build a Python function to check if a string: Starts and ends with double quotes (") Has at least one double quote somewhere inside (not including the first and last characters) 🧑‍💻 Function Implementation def within_and_has_double_quotes(s): # Check if string has at least 2 characters and starts/ends with a double quote if len(s) < 2 or s[0] != '"' or s[-1] != '"': return False # Check for another double quote in the interior (exclude first/last char) return '"' in s[1:-1] 🔍 Step-by-Step Explanation Check the boundary quotes: The string must be at least 2 characters. The first character is " and the last character is ". Check for an internal quote: Use s[1:-1] to get the inner part of the string (excluding first and last characters). Test if " exists in this interior slice. 🧪 Practice Questions Practice 1 print(within_and_has_double_quotes('"hello"')) # _______ Interior: hello (no quotes) Output: False Practice 2 print(within_and_has_double_quotes('"he"llo"')) # _______ Interior: he"llo Yes, has an interior double quote! Output: True Practice 3 print(within_and_has_double_quotes('""')) # _______ Interior: empty string "" Output: False Practice 4 print(within_and_has_double_quotes('"a"b"c"')) # _______ Interior: a"b"c Yes, contains double quotes inside! Output: True Practice 5 print(within_and_has_double_quotes('"no inner quote')) # _______ No ending double quote. Output: False ✅ Key Points Checks boundaries (" at both ends) Looks inside (middle of string) for at least one more double quote Returns True only if both conditions are met Happy Coding! 🚀

📝 Filter Words by Custom Criteria Below is the requested function to filter a list of words based on one of four criteria: continuous, vowel_rich, consonant_rich, or sorted. Words are checked case-insensitively, and helper functions are included for each criteria. def filter_words(words, criteria): """ Filters a list of words based on the given criteria. Args: words (list of str): Input list of words. criteria (str): One of 'continuous', 'vowel_rich', 'consonant_rich', 'sorted'. Returns: list or None: Filtered list based on criteria, or None if invalid. Criteria: • 'continuous': words ending with 'ing' (case-insensitive) • 'vowel_rich': words with more than 5 vowels • 'consonant_rich': words with more than 5 consonants • 'sorted': words whose letters are in ascending order (case-insensitive) """ def is_continuous(word): return word.lower().endswith('ing') def count_vowels(word): return sum(1 for ch in word.lower() if ch in "aeiou") def count_consonants(word): return sum(1 for ch in word.lower() if ch.isalpha() and ch not in "aeiou") def is_sorted(word): letters = [ch.lower() for ch in word if ch.isalpha()] return letters == sorted(letters) if criteria == 'continuous': return [w for w in words if is_continuous(w)] elif criteria == 'vowel_rich': return [w for w in words if count_vowels(w) > 5] elif criteria == 'consonant_rich': return [w for w in words if count_consonants(w) > 5] elif criteria == 'sorted': return [w for w in words if is_sorted(w)] else: return None ✨ Step-by-Step Explanation Helper Functions: is_continuous: Checks if the word ends with “ing” (case-insensitive). count_vowels: Counts vowels (a, e, i, o, u). count_consonants: Counts consonants (alphabetic letters that are not vowels). is_sorted: Checks if the letters of the word are in ascending order. Criteria Check: The function checks the criteria argument and applies the relevant helper. If an unknown criteria is given, returns None. 🧪 Practice Questions Try calling the function on the following list:

Check if a Number is Divisible by Exactly One of the Given NumbersWrite a function div_by_exactly_one that takes three integers num, a, and b. The function should return True if num is divisible by exactly one of the numbers a or b, and False otherwise. NOTE: This is a function type question, you don’t have to take input or print the output, just have to complete the required function definition.

def print_n_pattern(n): """ Prints an "N" shaped pattern with n rows. The pattern has vertical bars (|) on the left and right sides of each row, with a backslash (\) moving diagonally from the top-left to the bottom-right. There are no spaces to the right of the pattern. Args: n (int): The number of rows for the pattern (n > 0). """ for i in range(n): # Start each row with a vertical bar row = "|" # Add leading spaces before the diagonal backslash # The number of leading spaces is equal to the current row index 'i' row += " " * i # Add the diagonal backslash row += "\\" # Add trailing spaces after the diagonal backslash # The total width of the inner part (between vertical bars) is (n - 1) characters. # So, the number of trailing spaces is (n - 1) - i row += " " * (n - 1 - i) # End each row with a vertical bar row += "|" # Print the constructed row print(row) if __name__ == "__main__": try: # Read the integer input from the user n = int(input()) # Validate the input: n must be greater than 0 if n <= 0: print("Input 'n' must be a positive integer (n > 0).") else: # Call the function to print the pattern print_n_pattern(n) except ValueError: # Handle cases where the input is not a valid integer print("Invalid input. Please enter a single integer.")

✨ Print “X” Pattern Program Let’s write a Python program that prints an “X” shaped pattern using backslashes (\), forward slashes (/), and a lowercase x in the center! 😃 📋 Problem Description Input: An integer n (n >= 0) Output: An “X” shape with a center "x" and n identical rows above and below, built with slashes. No extra spaces to the right of the pattern. 🧑‍💻 Full Python Code n = int(input()) # Print the top part for i in range(n): # i spaces, then '\', then (2*n - 2*i - 1) spaces, then '/' left_spaces = ' ' * i middle_spaces = ' ' * (2 * n - 2 * i - 1) print(left_spaces + '\\' + middle_spaces + '/') # Print the central x print(' ' * n + 'x') # Print the bottom part (mirror of above) for i in range(n-1, -1, -1): left_spaces = ' ' * i middle_spaces = ' ' * (2 * n - 2 * i - 1) print(left_spaces + '/' + middle_spaces + '\\') 🔍 Step-by-Step Explanation Top Rows: For each row i from 0 to n-1: Print i spaces. Print a backslash (\). Print (2*n - 2*i - 1) spaces (between slashes). Print a forward slash (/). Central Row: Print n spaces, then the character 'x'. Bottom Rows: Repeat in reverse for i from n-1 to 0: Print i spaces. Print a forward slash (/). Print (2*n - 2*i - 1) spaces. Print a backslash (\). No trailing spaces are added to the right of the pattern. 🧪 Practice Examples Example 1 Input:

📝 Function: Replace Middle Element with n Copies Let’s build a helpful Python function that replaces the middle element of a tuple with n copies of itself! 🧑‍💻✨ 🚩 Function Definition def replace_middle_with_n_copies(t, n): # Find the middle index mid = len(t) // 2 # Create a tuple with 'n' copies of the middle element middle_replacement = (t[mid],) * n # Build the new tuple: before + replacement + after return t[:mid] + middle_replacement + t[mid+1:] 🔍 Step-by-Step Explanation Find the Middle Index Since the tuple has odd length, the middle index is len(t) // 2. Create Replacement (t[mid],) * n makes n copies of the middle element as a tuple. Assemble the Result Use slicing: t[:mid] (elements before the middle), middle_replacement, and t[mid+1:] (elements after the middle). 🤔 Practice Questions Practice 1 t = (1, 2, 3, 4, 5) n = 3 print(replace_middle_with_n_copies(t, n)) # What do you expect? Middle element: 3 Should be replaced by (3, 3, 3) Result: (1, 2, 3, 3, 3, 4, 5) Practice 2 t = ('a', 'b', 'c') n = 2 print(replace_middle_with_n_copies(t, n)) Middle element: 'b' Result: ('a', 'b', 'b', 'c') Practice 3 t = (10,) n = 4 print(replace_middle_with_n_copies(t, n)) Single element tuple: replace with 4 copies of itself. Result: (10, 10, 10, 10) ✨ Tips Tuples are immutable: You can’t change them; instead, create and return a new tuple. Odd length guaranteed: No need to check for even-length inputs. Works for any tuple type: numbers, strings, or mixed! Happy Python-ing! 🐍🚀