Set membership testing in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of set membership test with strings

What is the output of this Python code?

Python

fruits = {'apple', 'banana', 'cherry'}
print('banana' in fruits)
print('orange' in fruits)

A False\nTrue

B True\nTrue

C True\nFalse

D False\nFalse

💡 Hint

Check if each fruit is inside the set using the 'in' keyword.

The set contains 'banana' but not 'orange', so 'banana' in fruits is True and 'orange' in fruits is False.

❓ Predict Output

intermediate

Set membership with numbers and types

What will this code print?

Python

numbers = {1, 2, 3, 4}
print(3 in numbers)
print('3' in numbers)

A True\nTrue

B False\nFalse

C False\nTrue

D True\nFalse

💡 Hint

Remember that 3 (int) and '3' (string) are different types.

3 is in the set, but the string '3' is not, so the first print is True and the second is False.

❓ Predict Output

advanced

Set membership with mutable elements

What error does this code raise?

Python

my_set = {1, 2, [3, 4]}
print(2 in my_set)

A TypeError

B SyntaxError

C KeyError

D No error, prints True

💡 Hint

Sets cannot contain mutable elements like lists.

Lists are mutable and cannot be added to a set, so Python raises a TypeError.

❓ Predict Output

advanced

Set membership with nested tuples

What is the output of this code?

Python

nested_set = {(1, 2), (3, 4)}
print((1, 2) in nested_set)
print((2, 1) in nested_set)

A False\nTrue

B True\nFalse

C True\nTrue

D False\nFalse

💡 Hint

Tuples are ordered, so (1, 2) is different from (2, 1).

The tuple (1, 2) is in the set, but (2, 1) is not, so the output is True then False.

🧠 Conceptual

expert

Why does 'in' membership test run faster on sets?

Why is checking membership with 'in' faster on a set than on a list?

A Sets use hashing to check membership in constant time, while lists check each item one by one.

B Sets store elements in sorted order, so membership uses binary search.

C Sets convert elements to strings before checking membership, making it faster.

D Sets use multiple threads internally to check membership faster.

💡 Hint

Think about how sets organize data internally compared to lists.

Sets use a hash table which allows them to find elements quickly without checking every item, unlike lists which check each element in order.

Practice

1.

Which operator is used to check if an element exists inside a set in Python?

easy

A. contains

B. has

C. exists

D. in

Solution

1. Step 1: Understand set membership

   In Python, the in keyword checks if an element is present in a set.

2. Step 2: Identify correct operator

   Other options like has, exists, and contains are not valid Python operators for membership testing.

3. Final Answer:

   in -> Option D

4. Quick Check:

   Use in to test membership [OK]

Hint: Remember: use 'in' to check if item is inside a set [OK]

Common Mistakes:

• Using 'has' instead of 'in'
• Trying 'exists' keyword
• Using 'contains' which is not a Python operator

2.

Which of the following is the correct syntax to check if 5 is NOT in the set {1, 2, 3, 4}?

easy

A. 5 not in {1, 2, 3, 4}

B. not 5 in {1, 2, 3, 4}

C. 5 !in {1, 2, 3, 4}

D. 5 notinside {1, 2, 3, 4}

Solution

1. Step 1: Recall correct syntax for 'not in'

   The correct syntax to check if an element is not in a set is: element not in set.

2. Step 2: Evaluate each option

   5 not in {1, 2, 3, 4} uses correct syntax. not 5 in {1, 2, 3, 4} is incorrect because 'not' must come after the element. 5 !in {1, 2, 3, 4} uses invalid operator '!in'. 5 notinside {1, 2, 3, 4} uses a non-existent keyword 'notinside'.

3. Final Answer:

   5 not in {1, 2, 3, 4} -> Option A

4. Quick Check:

   Use 'not in' with element first [OK]

Hint: Write 'element not in set' to check absence [OK]

Common Mistakes:

• Using '!in' instead of 'not in'
• Placing 'not' before element
• Using invalid keywords like 'notinside'

3.

What will be the output of the following code?

my_set = {10, 20, 30}
print(25 in my_set)

medium

A. True

B. False

C. 25

D. Error

Solution

1. Step 1: Understand the set contents

   The set my_set contains 10, 20, and 30. It does not contain 25.

2. Step 2: Evaluate membership test

   The expression 25 in my_set checks if 25 is in the set. Since it is not, the result is False.

3. Final Answer:

   False -> Option B

4. Quick Check:

   25 not in set means False [OK]

Hint: If element missing in set, 'in' returns False [OK]

Common Mistakes:

• Assuming output is the element itself
• Confusing True/False for membership
• Expecting an error for missing element

4.

Find the error in this code snippet:

my_set = {1, 2, 3}
if 4 notin my_set:
    print("4 is not in the set")

medium

A. The keyword 'notin' is invalid

B. The set declaration is wrong

C. Missing colon after if statement

D. Print statement syntax error

Solution

1. Step 1: Check syntax of membership test

   The correct keyword to check absence is not in with a space, not notin.

2. Step 2: Verify other parts of code

   The set declaration is correct, colon after if is present, and print syntax is valid.

3. Final Answer:

   The keyword 'notin' is invalid -> Option A

4. Quick Check:

   Use 'not in' with space, not 'notin' [OK]

Hint: Remember: 'not in' has a space between words [OK]

Common Mistakes:

• Writing 'notin' as one word
• Forgetting colon after if
• Misreading print syntax

5.

Given the list nums = [1, 2, 2, 3, 4, 4, 5], which code snippet correctly prints Found only if 3 is in the unique set of numbers?

A) if 3 in nums:
       print("Found")

B) if 3 in set(nums):
       print("Found")

C) if 3 not in set(nums):
       print("Found")

D) if 3 not in nums:
       print("Found")

hard

A. if 3 in nums:
print("Found")

B. if 3 not in nums:
print("Found")

C. if 3 in set(nums):
print("Found")

D. if 3 not in set(nums):
print("Found")

Solution

1. Step 1: Understand the requirement

   We want to check if 3 is in the unique set of numbers from the list, so duplicates are ignored.

2. Step 2: Analyze each option

   if 3 in nums: print("Found") checks 3 in the list (with duplicates). if 3 in set(nums): print("Found") converts list to set and checks membership correctly. if 3 not in set(nums): print("Found") prints 'Found' if 3 is NOT in the set, which is wrong. if 3 not in nums: print("Found") checks 3 not in list, also wrong.

3. Final Answer:

   if 3 in set(nums): print("Found") -> Option C

4. Quick Check:

   Convert list to set before membership test [OK]

Hint: Convert list to set before membership test for uniqueness [OK]

Common Mistakes:

• Checking membership directly in list with duplicates
• Using 'not in' instead of 'in'
• Confusing list and set membership