Function - `sum_of_squares_of_even`

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🧮 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! 🚀

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: 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: 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! 🚀

✨ 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:

🔄 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:

📝 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: