Dictionary use cases in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of nested dictionary access

What is the output of this Python code?

Python

data = {'user': {'name': 'Alice', 'age': 30}, 'active': True}
print(data['user']['age'])

A 30

B 'Alice'

C True

D KeyError

💡 Hint

Look inside the nested dictionary under 'user' key.

The code accesses the 'user' dictionary inside 'data' and then gets the value for 'age', which is 30.

🧠 Conceptual

intermediate

Dictionary keys uniqueness

Which statement about Python dictionary keys is true?

A Dictionary keys can be duplicated if values differ.

B Dictionary keys must be unique and immutable.

C Dictionary keys can be mutable types like lists.

D Dictionary keys can be any data type including functions.

💡 Hint

Think about what happens if you try to use a list as a key.

Dictionary keys must be unique and immutable types like strings, numbers, or tuples. Mutable types like lists cannot be keys.

🔧 Debug

advanced

Identify the error in dictionary update

What error does this code raise?

Python

d = {'a': 1, 'b': 2}
d.update(['c', 3])

A No error, dictionary updated

B KeyError

C ValueError

D TypeError

💡 Hint

Check the argument type for update method.

The update method expects a mapping or iterable of key-value pairs. Passing a list of two elements causes a TypeError.

❓ Predict Output

advanced

Result of dictionary comprehension with condition

What is the output of this code?

Python

result = {x: x**2 for x in range(5) if x % 2 == 0}
print(result)

A {0: 0, 2: 4, 4: 16}

B {1: 1, 3: 9}

C SyntaxError

D {0: 0, 1: 1, 2: 4, 3: 9, 4: 16}

💡 Hint

Only even numbers are included as keys.

The comprehension includes only keys where x is even, so keys 0, 2, and 4 with their squares as values.

🚀 Application

expert

Count character frequency in string using dictionary

What is the value of 'freq' after running this code?

Python

text = 'banana'
freq = {}
for ch in text:
    freq[ch] = freq.get(ch, 0) + 1
print(freq)

A {'b': 1, 'a': 3, 'n': 2}

B {'b': 1, 'a': 2, 'n': 3}

C {'b': 1, 'a': 3, 'n': 3}

D KeyError

💡 Hint

Count how many times each letter appears in 'banana'.

The code counts each character's occurrences: 'b' once, 'a' three times, 'n' twice.

Practice

1. What is the main purpose of a dictionary in Python?

easy

A. To store data as key-value pairs for quick access

B. To store data in a fixed order like a list

C. To perform mathematical calculations

D. To create a sequence of numbers

Solution

1. Step 1: Understand dictionary structure

   Dictionaries store data in pairs where each key maps to a value.

2. Step 2: Identify dictionary use case

   This structure allows fast lookup by key, unlike lists which use indexes.

3. Final Answer:

   To store data as key-value pairs for quick access -> Option A

4. Quick Check:

   Dictionaries = key-value pairs [OK]

Hint: Remember: dictionaries use keys to find values fast [OK]

Common Mistakes:

• Confusing dictionaries with lists or sets
• Thinking dictionaries keep order like lists
• Assuming dictionaries store only numbers

2. Which of the following is the correct way to add a new key-value pair to a dictionary my_dict?

easy

A. my_dict.append('key', 'value')

B. my_dict.add('key', 'value')

C. my_dict['key'] = 'value'

D. my_dict.insert('key', 'value')

Solution

1. Step 1: Recall dictionary syntax for adding items

   To add or update a key-value pair, use square brackets with the key and assign a value.

2. Step 2: Check each option

   Only my_dict['key'] = 'value' correctly adds or updates the dictionary.

3. Final Answer:

   my_dict['key'] = 'value' -> Option C

4. Quick Check:

   Use square brackets to add/update dict items [OK]

Hint: Use square brackets and assignment to add dict items [OK]

Common Mistakes:

• Using list methods like append or insert on dictionaries
• Trying to use a non-existent add() method
• Confusing dictionary syntax with list syntax

3. What will be the output of the following code?

my_dict = {'a': 1, 'b': 2}
my_dict['c'] = 3
print(my_dict)

medium

A. Error: Cannot add new key

B. {'a': 1, 'b': 2}

C. {'c': 3}

D. {'a': 1, 'b': 2, 'c': 3}

Solution

1. Step 1: Understand dictionary update

   Adding my_dict['c'] = 3 adds a new key 'c' with value 3.

2. Step 2: Print the updated dictionary

   Printing shows all keys and values including the new one.

3. Final Answer:

   {'a': 1, 'b': 2, 'c': 3} -> Option D

4. Quick Check:

   Adding key updates dict content [OK]

Hint: Adding key-value pairs updates dictionary instantly [OK]

Common Mistakes:

• Expecting original dict without new key
• Thinking print shows only last added key
• Assuming error when adding new keys

4. Find the error in this code snippet:

my_dict = {'x': 10, 'y': 20}
print(my_dict['z'])

medium

A. KeyError because 'z' is not in the dictionary

B. SyntaxError due to wrong brackets

C. TypeError because dictionary keys must be integers

D. No error, prints 0

Solution

1. Step 1: Check dictionary keys

   Keys are 'x' and 'y', but 'z' is not present.

2. Step 2: Accessing missing key causes error

   Trying to print my_dict['z'] raises a KeyError.

3. Final Answer:

   KeyError because 'z' is not in the dictionary -> Option A

4. Quick Check:

   Access missing key = KeyError [OK]

Hint: Accessing missing keys causes KeyError [OK]

Common Mistakes:

• Assuming missing keys return 0 or None
• Confusing KeyError with SyntaxError
• Trying to use wrong bracket types

5. Given a list of tuples representing student names and scores:

students = [('Alice', 85), ('Bob', 90), ('Alice', 95)]

Which code correctly creates a dictionary with the latest score for each student?

hard

A. scores = dict(students)
scores['Alice'] = 85

B. scores = {name: score for name, score in students}

C. scores = {}
for name, score in students:
if name not in scores:
scores[name] = score

D. scores = {name: max(score) for name, score in students}

Solution

1. Step 1: Understand dictionary comprehension behavior

   When keys repeat, the last value overwrites earlier ones.

2. Step 2: Check each option

   scores = {name: score for name, score in students} uses comprehension that keeps the last score for 'Alice' (95). scores = dict(students) scores['Alice'] = 85 wrongly resets 'Alice' score. scores = {} for name, score in students: if name not in scores: scores[name] = score keeps only first score. scores = {name: max(score) for name, score in students} is invalid syntax.

3. Final Answer:

   scores = {name: score for name, score in students} -> Option B

4. Quick Check:

   Dict comprehension overwrites duplicate keys [OK]

Hint: Dict comprehension keeps last value for duplicate keys [OK]

Common Mistakes:

• Assuming dict keeps first value for duplicates
• Using invalid syntax for max(score)
• Manually overwriting values incorrectly