Python OOPE Exam

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🔄 Sum and Absolute Difference Alternately Let’s solve the problem step by step and make the code easy-to-follow with emojis, practice questions, and explanations! 🚀 📝 Full Python Code n = int(input()) for i in range(n): line = input().strip() # Split by comma and convert to integers a, b = map(int, line.split(',')) if i % 2 == 0: # Odd-numbered pair (1-based): print sum print(a + b) else: # Even-numbered pair (1-based): print absolute difference print(abs(a - b)) 🔍 Step-by-Step Explanation Read the number of pairs: The first input line is the integer n. Process each pair: For every line, read two comma-separated integers. Odd pair (1st, 3rd, 5th…): Print their sum. Even pair (2nd, 4th, 6th…): Print the absolute difference. Print the results one per line. 🧪 Practice Questions (with Solutions!) Practice 1 Input:

🔢 Function: Count Positive Integers (Ignore None) Let’s build an easy-to-use Python function! ✨ 🚩 Function Definition def count_positive_integers(lst): return sum(1 for x in lst if x is not None and isinstance(x, int) and x > 0) 📝 Step-by-Step Explanation Iterate through each element x in the list. Ignore None values (x is not None). Check if the value is an integer (isinstance(x, int)). Count only if the integer is positive (x > 0, so zero is ignored). Sum up the count (sum(1 for ...)). 🧪 Practice Questions Practice 1 lst = [4, -3, None, 7, 0, None, 10] print(count_positive_integers(lst)) # _______ Answer: 3 (Counts 4, 7, 10)

🚀 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: 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! 🐍💡

📝 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:

✨ 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: