Array comparison and set operations in Ruby - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Ruby code using array intersection?

Consider the following Ruby code snippet. What will it print?

Ruby

a = [1, 2, 3, 4]
b = [3, 4, 5, 6]
puts (a & b).inspect

A [5, 6]

B [1, 2, 3, 4, 5, 6]

C [3, 4]

D [1, 2]

💡 Hint

The & operator returns elements common to both arrays.

The & operator in Ruby returns a new array containing elements common to both arrays, without duplicates. Here, 3 and 4 are common.

❓ Predict Output

intermediate

What does this Ruby code print when using array difference?

Look at this Ruby code. What is the output?

Ruby

a = [1, 2, 3, 4]
b = [3, 4, 5, 6]
puts (a - b).inspect

A [1, 2]

B [3, 4]

C [1, 2, 3, 4, 5, 6]

D [5, 6]

💡 Hint

The - operator removes elements from the first array that appear in the second.

The - operator returns a new array with elements from the first array that are not in the second. So 3 and 4 are removed from a.

❓ Predict Output

advanced

What is the output of this Ruby code using array union?

What will this Ruby code print?

Ruby

a = [1, 2, 3]
b = [3, 4, 5]
puts (a | b).inspect

A [3, 4, 5]

B [1, 2, 3, 4, 5]

C [1, 2]

D [1, 2, 3, 3, 4, 5]

💡 Hint

The | operator returns all unique elements from both arrays.

The | operator returns a new array with unique elements from both arrays combined. Duplicates are removed.

❓ Predict Output

advanced

What error does this Ruby code raise when comparing arrays with a non-array?

What error will this Ruby code produce?

Ruby

a = [1, 2, 3]
b = 5
puts (b & a).inspect

A ArgumentError: wrong number of arguments (given 1, expected 0)

B TypeError: no implicit conversion of Integer into Enumerator

C TypeError: no implicit conversion of Integer into Array

D NoMethodError: undefined method `&' for Integer

💡 Hint

The & operator expects both operands to be arrays.

The & operator is defined for arrays. Using it with an Integer on the left causes NoMethodError because Integer does not have & method.

🧠 Conceptual

expert

How many unique elements are in the result of this Ruby array operation?

Given the arrays below, how many elements are in the array resulting from (a | b) - (a & b)?

Ruby

a = [1, 2, 3, 4, 5]
b = [4, 5, 6, 7]
result = (a | b) - (a & b)
puts result.size

A 5

B 4

C 6

D 7

💡 Hint

Think about the symmetric difference: elements in either array but not both.

The expression (a | b) - (a & b) gives the symmetric difference. Here, a | b is [1,2,3,4,5,6,7], a & b is [4,5]. Removing [4,5] leaves [1,2,3,6,7], which has 5 elements.

Practice

1.

Which operator in Ruby returns the common elements between two arrays?

[1, 2, 3] ? [2, 3, 4]

easy

A. & (intersection)

B. | (union)

C. - (difference)

D. == (equality)

Solution

1. Step 1: Understand the operators

   The & operator returns elements common to both arrays.

2. Step 2: Apply to example arrays

   For arrays [1, 2, 3] and [2, 3, 4], common elements are 2 and 3.

3. Final Answer:

   & (intersection) -> Option A

4. Quick Check:

   Common elements = & [OK]

Hint: Use & for common items between arrays [OK]

Common Mistakes:

• Confusing union (|) with intersection (&)
• Using - for common elements
• Thinking == compares elements regardless of order

2.

Which of the following is the correct syntax to find the difference between two arrays a and b in Ruby?

easy

A. a == b

B. a - b

C. a | b

D. a & b

Solution

1. Step 1: Recall difference operator

   The - operator returns elements in a not in b.

2. Step 2: Confirm syntax correctness

   Using a - b is the correct Ruby syntax for difference.

3. Final Answer:

   a - b -> Option B

4. Quick Check:

   Difference = - [OK]

Hint: Use a - b to get items in a but not in b [OK]

Common Mistakes:

• Using & instead of - for difference
• Using | which is union
• Using == which checks equality

3.

What is the output of the following Ruby code?

a = [1, 2, 3, 4]
b = [3, 4, 5, 6]
result = a | b
puts result.inspect

medium

A. [1, 2, 3, 4, 5, 6]

B. [3, 4]

C. [1, 2]

D. [1, 2, 3, 4]

Solution

1. Step 1: Understand the union operator

   The | operator combines all unique elements from both arrays.

2. Step 2: Apply union to arrays

   Arrays a and b combined uniquely are [1, 2, 3, 4, 5, 6].

3. Final Answer:

   [1, 2, 3, 4, 5, 6] -> Option A

4. Quick Check:

   Union = | [OK]

Hint: Use | to combine unique elements from both arrays [OK]

Common Mistakes:

• Confusing union with intersection
• Expecting duplicates in output
• Using - which removes elements

4.

Identify the error in the following Ruby code snippet:

a = [1, 2, 3]
b = [2, 3, 4]
result = a & b
puts result

What is the problem?

medium

A. Syntax error due to missing parentheses

B. Outputs all elements from both arrays

C. Runtime error because & is not defined for arrays

D. No error; code runs and outputs common elements

Solution

1. Step 1: Check syntax and operators

   The & operator is valid for arrays to find intersection.

2. Step 2: Verify output behavior

   The code outputs common elements [2, 3] without error.

3. Final Answer:

   No error; code runs and outputs common elements -> Option D

4. Quick Check:

   Intersection & works fine [OK]

Hint: & works for arrays to find common items without syntax errors [OK]

Common Mistakes:

• Thinking & causes syntax error
• Expecting union instead of intersection
• Confusing output with difference

5.

You have two arrays representing user IDs:

active_users = [101, 102, 103, 104, 105]
inactive_users = [103, 105, 107]

Which Ruby expression gives you the list of users who are active but not inactive?

hard

A. active_users | inactive_users

B. active_users & inactive_users

C. active_users - inactive_users

D. active_users == inactive_users

Solution

1. Step 1: Understand the problem

   We want users in active_users but not in inactive_users.

2. Step 2: Use difference operator

   The - operator returns elements in first array not in second.

3. Step 3: Apply to given arrays

   active_users - inactive_users = [101, 102, 104]

4. Final Answer:

   active_users - inactive_users -> Option C

5. Quick Check:

   Difference = - [OK]

Hint: Use - to exclude inactive from active users [OK]

Common Mistakes:

• Using & which gives only common users
• Using | which combines all users
• Using == which checks equality