Python OOPE Exam

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📝 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! 🐍🚀

🔄 Swap and Reverse Lines Program in Python Below is a Python program that reads multiple lines of input, swaps every two consecutive lines, and then reverses every odd-numbered line in the resulting output sequence. This meets the problem’s requirements exactly. 🛠️ Complete Code def main(): n = int(input()) lines = [input() for _ in range(n)] # Swap every two consecutive lines for i in range(0, n - 1, 2): lines[i], lines[i + 1] = lines[i + 1], lines[i] # Reverse every odd-numbered line in the output (1-based index) for i in range(n): if (i + 1) % 2 == 1: # odd line number lines[i] = lines[i][::-1] # Print the transformed lines for line in lines: print(line) if __name__ == "__main__": main() 📝 Explanation Input Reading: Read the integer n for the number of lines, then read each line into a list. Swapping Consecutive Lines: Iterate through the list in steps of two, swapping the pair at indices i and i+1. Reversing Odd-Numbered Lines: After swapping, reverse every odd line (1st, 3rd, 5th, …) by checking (i+1) % 2 == 1. Output: Print each line after these transformations. 🌟 Example Run Input:

📝 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.

✨ 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: Find Last Word Starting with Uppercase Here’s how you can implement the last_word_starts_with_upper_case function to find the last word in a sentence that starts with an uppercase letter. If no such word exists, it returns None. 🛠️ Function Definition def last_word_starts_with_upper_case(sentence: str): """ Find the last word in a sentence that starts with an uppercase letter. Args: sentence (str): The input sentence. Returns: str or None: The last word starting with an uppercase letter, or None if no such word exists. """ result = None for word in sentence.split(): if word and word[0].isupper(): result = word return result ✨ Step-by-Step Explanation Step 1: Split the sentence into individual words using split(). Step 2: Loop through each word: If the word starts with an uppercase letter (word.isupper()), save it as the current result. Step 3: After checking all words, return the last word found that met the criteria, or None if none did. 🚦 Example Usage print(last_word_starts_with_upper_case("This is a Test sentence")) # Output: "Test" print(last_word_starts_with_upper_case("no uppercase words here")) # Output: None 📝 Practice Questions What will last_word_starts_with_upper_case("Alice and Bob went to Paris") return? Try last_word_starts_with_upper_case("all lowercase words"). What is the result of last_word_starts_with_upper_case("Python Is Easy To Learn")? ✅ Solutions "Paris" None "Learn" With this function, you can quickly scan sentences for the last word that begins with a capital letter. Perfect for text analysis and simple parsing tasks! 🐍💡