Array comparison and set operations in Ruby - Step-by-Step Execution

Concept Flow - Array comparison and set operations

Start with two arrays

↓

Compare arrays for equality

↓

Perform set operations

↓

Union

↓

Result output

Start with two arrays, compare them, then perform union, intersection, and difference operations to get new arrays.

Execution Sample

Ruby

a = [1, 2, 3, 4]
b = [3, 4, 5, 6]

puts a == b
puts (a | b).inspect
puts (a & b).inspect
puts (a - b).inspect

This code compares two arrays for equality and shows their union, intersection, and difference.

Execution Table

Step	Action	Expression	Result	Explanation
1	Compare arrays a and b	a == b	false	Arrays differ in elements, so equality is false
2	Union of a and b	a | b	[1, 2, 3, 4, 5, 6]	Combines unique elements from both arrays
3	Intersection of a and b	a & b	[3, 4]	Elements common to both arrays
4	Difference of a and b	a - b	[1, 2]	Elements in a not in b
5	End	-	-	All operations completed

💡 All set operations performed and results displayed

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	Final
a	[1, 2, 3, 4]	[1, 2, 3, 4]	[1, 2, 3, 4]	[1, 2, 3, 4]	[1, 2, 3, 4]	[1, 2, 3, 4]
b	[3, 4, 5, 6]	[3, 4, 5, 6]	[3, 4, 5, 6]	[3, 4, 5, 6]	[3, 4, 5, 6]	[3, 4, 5, 6]
a == b	N/A	false	false	false	false	false
a | b	N/A	N/A	[1, 2, 3, 4, 5, 6]	[1, 2, 3, 4, 5, 6]	[1, 2, 3, 4, 5, 6]	[1, 2, 3, 4, 5, 6]
a & b	N/A	N/A	N/A	[3, 4]	[3, 4]	[3, 4]
a - b	N/A	N/A	N/A	N/A	[1, 2]	[1, 2]

Key Moments - 3 Insights

Why does 'a == b' return false even though some elements are the same?

Because '==' checks if arrays have the same elements in the same order. Here, a and b differ in elements and order, so result is false (see step 1 in execution_table).

What does the '|' operator do with arrays?

It creates a union of unique elements from both arrays, removing duplicates (see step 2 in execution_table).

Why does 'a - b' only show elements from a not in b?

Because '-' subtracts elements of b from a, leaving only those unique to a (see step 4 in execution_table).

Visual Quiz - 3 Questions

Look at the execution table, what is the result of 'a & b' at step 3?

A [1, 2]

B [3, 4]

C [5, 6]

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

💡 Hint

Check the 'Result' column for step 3 in execution_table

At which step does the union of arrays a and b get computed?

A Step 1

B Step 3

C Step 2

D Step 4

💡 Hint

Look for the action 'Union of a and b' in execution_table

If array b was changed to [1, 2, 3, 4], what would 'a == b' return?

A true

B false

C nil

D error

💡 Hint

Refer to variable_tracker and key_moments about array equality

Concept Snapshot

Array comparison and set operations in Ruby:
- Use '==' to check if arrays are equal (same elements, same order)
- Use '|' for union (unique elements from both arrays)
- Use '&' for intersection (common elements)
- Use '-' for difference (elements in first array not in second)
- Results are new arrays without modifying originals

Full Transcript

This visual execution trace shows how Ruby compares two arrays and performs set operations. First, it checks if arrays a and b are equal using '==', which returns false because their elements differ. Then it computes the union with '|', combining unique elements from both arrays. Next, it finds the intersection with '&', showing elements common to both. Finally, it calculates the difference with '-', showing elements in a not in b. Variables remain unchanged throughout. Key moments clarify why equality requires exact match and how set operators work. The quiz tests understanding of these steps.

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