Python OOPE Exam

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🏃‍♂️ Matrix Walk Function Let’s design a Python function that follows three special matrix paths — “L”, “Z”, and “O” shapes — and collects the matrix elements accordingly! 🚀 🧑‍💻 Function Definition def matrix_walk(matrix, path): n = len(matrix) result = [] if path == "L": # Down the first column for i in range(n): result.append(matrix[i][0]) # Across the bottom row, excluding first element for j in range(1, n): result.append(matrix[n-1][j]) return result elif path == "Z": # Top row for j in range(n): result.append(matrix[0][j]) # Diagonal except first and last row for i in range(1, n-1): result.append(matrix[i][n-i-1]) # Bottom row for j in range(n): result.append(matrix[n-1][j]) return result elif path == "O": # Top row for j in range(n): result.append(matrix[0][j]) # Right column (excluding top and bottom) for i in range(1, n-1): result.append(matrix[i][n-1]) # Bottom row (reverse, excluding last element) for j in range(n-1, -1, -1): result.append(matrix[n-1][j]) # Left column (reverse, excluding top and bottom) for i in range(n-2, 0, -1): result.append(matrix[i][0]) return result else: return [] 🔍 Step-by-Step Explanation 1. L-Shape (“L”) Walk down the first column (leftmost). Then walk right along the bottom row (skipping the already-added first element). 2. Z-Shape (“Z”) Walk across the top row (left to right). Walk diagonally from the top-right to bottom-left, excluding corners. Walk across the bottom row (left to right). 3. O-Shape (“O”) Walk around the outside (clockwise): Top row (left to right), Right column (top to bottom, skipping corners), Bottom row (right to left), Left column (bottom to top, skipping corners). 🧪 Practice Questions with Solutions Example Matrix matrix = [ [1, 2, 3], [4, 5, 6], [7, 8, 9] ] Practice 1: L-Shape print(matrix_walk(matrix, "L")) # Output: [1, 4, 7, 8, 9] Practice 2: Z-Shape print(matrix_walk(matrix, "Z")) # Output: [1, 2, 3, 5, 7, 8, 9] Practice 3: O-Shape print(matrix_walk(matrix, "O")) # Output: [1, 2, 3, 6, 9, 8, 7, 4] ✨ Key Points The function traverses only as per the specified path — “L”, “Z”, or “O”. For unknown paths, returns an empty list. Works for any square matrix of size 1 or more. Happy Matrix Walking! 🟦🟩🟨🟥

🧮 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! 🚀

✨ 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:

📝 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! 🚀

📝 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! 🐍🚀