Module 8: Practice Problems

Problem 1: Create a Dictionary Easy

Create a dictionary called book with the keys "title", "author", and "year". Assign appropriate values. Print each value on its own line.

book = {
    "title": "To Kill a Mockingbird",
    "author": "Harper Lee",
    "year": 1960
}
print(book["title"])
print(book["author"])
print(book["year"])

Problem 2: Safe Access with get() Easy

Given the dictionary person = {"name": "Alice", "age": 30}, use get() to print the "email" value with a default of "Not provided".

person = {"name": "Alice", "age": 30}
email = person.get("email", "Not provided")
print(email)  # Not provided

Problem 3: Add and Update Easy

Start with scores = {"math": 85, "science": 90}. Add a new key "english" with value 88, then update "math" to 92. Print the dictionary.

scores = {"math": 85, "science": 90}
scores["english"] = 88
scores["math"] = 92
print(scores)
# {'math': 92, 'science': 90, 'english': 88}

Problem 4: Loop Through a Dictionary Easy

Given ages = {"Alice": 25, "Bob": 30, "Carol": 28}, loop through the dictionary and print each name and age in the format "Name is X years old".

ages = {"Alice": 25, "Bob": 30, "Carol": 28}
for name, age in ages.items():
    print(f"{name} is {age} years old")

Problem 5: Create and Unpack Tuples Easy

Create a tuple called point with values (5, 10). Unpack it into variables x and y, then print both.

point = (5, 10)
x, y = point
print(f"x = {x}, y = {y}")

Problem 6: Word Counter Medium

Write code that counts how many times each word appears in the string "the cat and the dog and the bird". Store the counts in a dictionary and print it.

text = "the cat and the dog and the bird"
counts = {}
for word in text.split():
    if word in counts:
        counts[word] += 1
    else:
        counts[word] = 1
print(counts)
# {'the': 3, 'cat': 1, 'and': 2, 'dog': 1, 'bird': 1}

Problem 7: Merge Two Dictionaries Medium

Given defaults = {"color": "blue", "size": 12, "font": "Arial"} and user_prefs = {"color": "red", "bold": True}, merge user_prefs into defaults and print the result.

defaults = {"color": "blue", "size": 12, "font": "Arial"}
user_prefs = {"color": "red", "bold": True}
defaults.update(user_prefs)
print(defaults)
# {'color': 'red', 'size': 12, 'font': 'Arial', 'bold': True}

Problem 8: Nested Dictionary Access Medium

Given the nested dictionary below, print the GPA of the student named Bob.

school = {
    "alice": {"age": 20, "gpa": 3.9},
    "bob": {"age": 21, "gpa": 3.4}
}

print(school["bob"]["gpa"])  # 3.4

Problem 9: Tuple as Dictionary Key Medium

Create a dictionary that maps coordinate tuples to city names: (40.7, -74.0) to "New York" and (51.5, -0.1) to "London". Look up and print the city at (51.5, -0.1).

cities = {
    (40.7, -74.0): "New York",
    (51.5, -0.1): "London"
}
print(cities[(51.5, -0.1)])  # London

Problem 10: Find the Max Value Medium

Given prices = {"apple": 1.20, "banana": 0.50, "cherry": 3.00, "date": 5.50}, find and print the most expensive fruit and its price.

prices = {"apple": 1.20, "banana": 0.50, "cherry": 3.00, "date": 5.50}

max_fruit = ""
max_price = 0

for fruit, price in prices.items():
    if price > max_price:
        max_price = price
        max_fruit = fruit

print(f"Most expensive: {max_fruit} at ${max_price:.2f}")
# Most expensive: date at $5.50

Problem 11: Swap Variables with Tuples Easy

Given a = 100 and b = 200, swap their values using tuple unpacking (without a temporary variable). Print both values before and after the swap.

a = 100
b = 200
print(f"Before: a={a}, b={b}")
a, b = b, a
print(f"After: a={a}, b={b}")
# Before: a=100, b=200
# After: a=200, b=100

Problem 12: Invert a Dictionary Hard

Given grades = {"Alice": "A", "Bob": "B", "Carol": "A", "Dave": "C"}, create a new dictionary that maps each grade to a list of students with that grade.

grades = {"Alice": "A", "Bob": "B", "Carol": "A", "Dave": "C"}

by_grade = {}
for student, grade in grades.items():
    if grade not in by_grade:
        by_grade[grade] = []
    by_grade[grade].append(student)

print(by_grade)
# {'A': ['Alice', 'Carol'], 'B': ['Bob'], 'C': ['Dave']}

Problem 13: Contact Book Hard

Build a simple contact book as a dictionary of dictionaries. Each contact should have a name (as the key), and a nested dictionary with "phone" and "email". Add at least 3 contacts, then loop through and print each contact's information.

contacts = {
    "Alice": {"phone": "555-0101", "email": "alice@example.com"},
    "Bob": {"phone": "555-0102", "email": "bob@example.com"},
    "Carol": {"phone": "555-0103", "email": "carol@example.com"}
}

for name, info in contacts.items():
    print(f"{name}: {info['phone']} | {info['email']}")

Problem 14: Student Averages Hard

Given the dictionary below, calculate and print each student's average score:

students = {
    "Alice": (90, 85, 92),
    "Bob": (78, 82, 80),
    "Carol": (95, 98, 97)
}

students = {
    "Alice": (90, 85, 92),
    "Bob": (78, 82, 80),
    "Carol": (95, 98, 97)
}

for name, scores in students.items():
    avg = sum(scores) / len(scores)
    print(f"{name}: {avg:.1f}")
# Alice: 89.0
# Bob: 80.0
# Carol: 96.7

Problem 15: Choose the Right Structure Medium

For each scenario, state whether you would use a list, tuple, or dictionary, and explain why:

1. Storing RGB color values (red=255, green=128, blue=0)
2. A shopping cart where items can be added and removed
3. Mapping country codes to country names
4. The days of the week
5. Tracking the number of times each letter appears in a word