Python OOPE Exam

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📝 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:

🚀 Function: increment_value_with_max_limit Here’s a function to increment the value for a given key in a dictionary, making sure it never exceeds a specified maximum limit! def increment_value_with_max_limit(d, key, inc, limit): """ Increment the value of d[key] by inc, but do not let it exceed limit. Args: d (dict): Dictionary with integer values. key: Key whose value you want to increment. inc (int): The increment value. limit (int): The maximum allowed value. Returns: None: The dictionary is modified in-place. Example: d = {'a': 5} increment_value_with_max_limit(d, 'a', 10, 12) # Now d['a'] == 12 """ if key in d: d[key] = min(d[key] + inc, limit) ✨ Step-by-Step Explanation Check Key Exists: Make sure the key is present in the dictionary. Increment and Cap: Add the inc value to the current value for the key. Use min(new_value, limit) to cap the result at limit if it would go over. Update In Place: The original dictionary is modified directly. No return value necessary! 📝 Practice Questions If d = {'x': 7}, what does increment_value_with_max_limit(d, 'x', 4, 10) do? What happens for d = {'b': 2}, increment_value_with_max_limit(d, 'b', 5, 5)? Try d = {'score': 20}, increment_value_with_max_limit(d, 'score', 3, 22) ✅ Solutions The value becomes min(7+4, 10) = 10. So, d['x'] will be 10. The value becomes min(2+5, 5) = 5. So, d['b'] will be 5. The value becomes min(20+3, 22) = 22. So, d['score'] will be 22. 💡 Now you can safely increment dictionary values without ever exceeding the maximum you set!

📝 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:

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.")

🧮 Function: is_positive_odd_or_negative_even Let’s build a Python function that checks if a number is either a positive odd or a negative even. 🎯 🚩 Function Definition def is_positive_odd_or_negative_even(n): # Check for positive odd OR negative even return (n > 0 and n % 2 == 1) or (n < 0 and n % 2 == 0) 📝 Step-by-Step Explanation Positive Odd Number: Condition: n > 0 and n % 2 == 1 Checks if the number is positive AND odd. Negative Even Number: Condition: n < 0 and n % 2 == 0 Checks if the number is negative AND even. Logical OR: The function returns True if any one of the above conditions is met. 🧪 Practice Questions Practice 1 print(is_positive_odd_or_negative_even(7)) # _______ 7 is positive and odd. Output: True Practice 2 print(is_positive_odd_or_negative_even(-4)) # _______ -4 is negative and even. Output: True Practice 3 print(is_positive_odd_or_negative_even(-3)) # _______ -3 is negative and odd. Output: False Practice 4 print(is_positive_odd_or_negative_even(8)) # _______ 8 is positive and even. Output: False Practice 5 print(is_positive_odd_or_negative_even(0)) # _______ 0 is neither positive nor negative (and it’s even). Output: False ✨ Key Points Odd number: Remainder when divided by 2 is 1 (n % 2 == 1). Even number: Remainder when divided by 2 is 0 (n % 2 == 0). Positive Odd or Negative Even: Only one of these is needed for True. Zero case: 0 is not positive or negative, so always returns False. Happy Coding! 🚀