Subset and superset checks in Python - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if set A is a subset of set B.

Python

A = {1, 2, 3}
B = {1, 2, 3, 4, 5}
result = A.[1](B)
print(result)

Drag options to blanks, or click blank then click option'

Aissubset

Bissuperset

Cunion

Dintersection

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'issuperset' instead of 'issubset'.

Using 'union' or 'intersection' which return new sets, not booleans.

The issubset method checks if all elements of set A are in set B.

2fill in blank

medium

Complete the code to check if set B is a superset of set A.

Python

A = {1, 2}
B = {1, 2, 3, 4}
result = B.[1](A)
print(result)

Drag options to blanks, or click blank then click option'

Aissuperset

Bissubset

Cdifference

Dsymmetric_difference

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'issubset' instead of 'issuperset'.

Using set operations that return sets instead of booleans.

The issuperset method checks if set B contains all elements of set A.

3fill in blank

hard

Fix the error in the code to check if set A is a subset of set B.

Python

A = {5, 6}
B = {4, 5, 6, 7}
result = A.[1](B)
print(result)

Drag options to blanks, or click blank then click option'

Aissuperset

Bissuper

Cissubset

Dsubset

Attempts:

3 left

💡 Hint

Common Mistakes

Using non-existent methods like 'issuper' or 'subset'.

Confusing 'issubset' with 'issuperset'.

The correct method to check if A is a subset of B is issubset. Other options are invalid or incorrect.

4fill in blank

hard

Fill both blanks to create a dictionary of words and their lengths, including only words longer than 3 letters.

Python

words = ['apple', 'bat', 'carrot', 'dog']
lengths = {word: [1] for word in words if len(word) [2] 3}
print(lengths)

Drag options to blanks, or click blank then click option'

Alen(word)

B>

C<

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself instead of its length.

Using '<' instead of '>' causing wrong filtering.

We use len(word) to get the length, and filter words with length greater than 3 using >.

5fill in blank

hard

Fill all three blanks to create a dictionary with uppercase keys and values only if the value is greater than 0.

Python

data = {'a': 1, 'b': 0, 'c': 3}
result = { [1]: [2] for k, v in data.items() if v [3] 0 }
print(result)

Drag options to blanks, or click blank then click option'

Ak.upper()

Bv

C>

Dk

Attempts:

3 left

💡 Hint

Common Mistakes

Using original keys instead of uppercase keys.

Filtering with '<' instead of '>'.

Using keys as values by mistake.

The keys are converted to uppercase with k.upper(), values are v, and we filter values greater than 0 using >.

Practice

1. Which of the following statements correctly describes a subset in Python sets?

easy

A. All elements of set A are in set B.

B. Set A has more elements than set B.

C. Set A and set B have no common elements.

D. Set A and set B have exactly the same elements.

Solution

1. Step 1: Understand subset definition

   A set A is a subset of set B if every element in A is also in B.

2. Step 2: Match definition to options

   All elements of set A are in set B. states all elements of A are in B, which matches the subset definition.

3. Final Answer:

   All elements of set A are in set B. -> Option A

4. Quick Check:

   Subset means all elements inside another set [OK]

Hint: Subset means every item of one set is inside another [OK]

Common Mistakes:

• Confusing subset with superset
• Thinking subsets must have fewer elements
• Assuming no common elements means subset

2. Which of the following is the correct syntax to check if set A is a superset of set B in Python?

easy

A. A.issubset(B)

B. A <= B

C. A.issuperset(B)

D. A < B

Solution

1. Step 1: Recall superset method

   To check if A is a superset of B, use A.issuperset(B) or A >= B.

2. Step 2: Identify correct option

   A.issuperset(B) uses A.issuperset(B), which is the correct method.

3. Final Answer:

   A.issuperset(B) -> Option C

4. Quick Check:

   Superset check uses issuperset() method [OK]

Hint: Use issuperset() to check if A contains all of B [OK]

Common Mistakes:

• Using issubset() instead of issuperset()
• Using <= or < for superset checks
• Confusing method names

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

set_a = {1, 2, 3}
set_b = {1, 2, 3, 4, 5}
print(set_a <= set_b)
print(set_b >= set_a)

medium

A. False\nFalse

B. False\nTrue

C. True\nFalse

D. True\nTrue

Solution

1. Step 1: Evaluate set_a <= set_b

   set_a has elements {1,2,3}, all of which are in set_b, so set_a is subset of set_b, result is True.

2. Step 2: Evaluate set_b >= set_a

   set_b contains all elements of set_a, so set_b is superset of set_a, result is True.

3. Final Answer:

   True True -> Option D

4. Quick Check:

   Subset and superset checks both True [OK]

Hint: <= and >= check subset and superset respectively [OK]

Common Mistakes:

• Mixing up <= and >= operators
• Assuming strict subset/superset without equality
• Forgetting sets can be equal

4. The following code is intended to check if set A is a subset of set B, but it raises an error. What is the problem?

set_a = {1, 2}
set_b = {1, 2, 3}
result = set_a.subset(set_b)
print(result)

medium

A. Sets cannot be compared using methods.

B. The method name should be issubset(), not subset().

C. The sets must be converted to lists first.

D. The print statement is missing parentheses.

Solution

1. Step 1: Identify method name error

   The correct method to check subset is issubset(), not subset().

2. Step 2: Confirm correct usage

   Using set_a.issubset(set_b) returns True or False without error.

3. Final Answer:

   The method name should be issubset(), not subset(). -> Option B

4. Quick Check:

   issubset() is correct method name [OK]

Hint: Use issubset(), not subset(), to check subsets [OK]

Common Mistakes:

• Using wrong method names
• Thinking sets need conversion to lists
• Confusing print syntax errors

5. Given two sets:

set_x = {2, 4, 6, 8}
set_y = {4, 6}

Which of the following expressions will return True for checking if set_y is a proper subset of set_x (subset but not equal)?

hard

A. set_y < set_x

B. set_y >= set_x

C. set_x.issubset(set_y)

D. set_y.issuperset(set_x)

Solution

1. Step 1: Understand proper subset

   A proper subset means all elements of set_y are in set_x, and sets are not equal.

2. Step 2: Analyze options

   set_y < set_x uses < operator which checks proper subset (subset but not equal). set_y >= set_x checks superset (wrong direction). set_x.issubset(set_y) reverses the sets (wrong). set_y.issuperset(set_x) is wrong (issuperset).

3. Final Answer:

   set_y < set_x -> Option A

4. Quick Check:

   Proper subset uses < operator [OK]

Hint: Use < for proper subset (subset but not equal) [OK]

Common Mistakes:

• Using >= which checks superset
• Reversing the subset direction
• Confusing issubset() with issuperset()