Array slicing and ranges in Ruby - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Ruby code using range slicing?

Consider the following Ruby code snippet. What will be printed?

Ruby

arr = [10, 20, 30, 40, 50]
puts arr[1..3].inspect

A [20, 30, 40, 50]

B [20, 30, 40]

C [10, 20, 30]

D [30, 40, 50]

💡 Hint

Remember that the range 1..3 includes elements at index 1, 2, and 3.

The range 1..3 selects elements starting at index 1 up to and including index 3. So arr[1..3] returns [20, 30, 40].

❓ Predict Output

intermediate

What does this code output when slicing with an exclusive range?

Look at this Ruby code. What will it print?

Ruby

arr = ['a', 'b', 'c', 'd', 'e']
puts arr[0...3].inspect

A ["b", "c", "d"]

B ["a", "b", "c", "d"]

C ["a", "b", "c"]

D ["a", "b"]

💡 Hint

The three dots in the range exclude the last index.

The range 0...3 includes indices 0, 1, and 2 but excludes 3. So arr[0...3] returns ["a", "b", "c"].

❓ Predict Output

advanced

What is the output of this code using negative indices in slicing?

Analyze this Ruby code and determine its output.

Ruby

arr = [1, 2, 3, 4, 5, 6]
puts arr[-4..-2].inspect

A [3, 4, 5]

B [4, 5, 6]

C [2, 3, 4]

D [3, 4]

💡 Hint

Negative indices count from the end: -1 is last element.

Index -4 corresponds to element 3, -2 corresponds to element 5. The range -4..-2 includes indices -4, -3, -2, so elements [3, 4, 5].

❓ Predict Output

advanced

What error does this Ruby code raise when slicing with an invalid range?

What happens when you run this Ruby code?

Ruby

arr = [7, 8, 9]
puts arr[5..7].inspect

A ArgumentError

B nil

C IndexError

D []

💡 Hint

Check what Ruby returns when slicing beyond array length.

Ruby returns nil when the start index of the range is beyond the array length. So arr[5..7] returns nil, not an error or empty array.

🧠 Conceptual

expert

How many elements are in the resulting array after this slicing?

Given this Ruby code, how many elements does the resulting array have?

Ruby

arr = (1..10).to_a
result = arr[2..8]
puts result.size

A 7

B 6

C 8

D 9

💡 Hint

Remember that the range 2..8 includes both ends.

The slice arr[2..8] includes elements at indices 2 through 8, which is 7 elements total.

Practice

1. What does the expression arr[1..3] do when arr = [10, 20, 30, 40, 50]?

easy

A. Returns a new array with elements at indexes 1, 2, and 3

B. Returns a new array with elements at indexes 1 and 3 only

C. Returns a new array with elements at indexes 1 and 2 only

D. Returns a single element at index 3

Solution

1. Step 1: Understand the range operator '..'

   The range 1..3 includes both 1 and 3, so it covers indexes 1, 2, and 3.

2. Step 2: Apply slicing to the array

   Slicing arr[1..3] returns elements at indexes 1, 2, and 3, which are 20, 30, and 40.

3. Final Answer:

   Returns a new array with elements at indexes 1, 2, and 3 -> Option A

4. Quick Check:

   Range '..' includes end index = elements 20, 30, 40 [OK]

Hint: Remember '..' includes the end index in slicing [OK]

Common Mistakes:

• Thinking '..' excludes the end index
• Confusing indexes with values
• Assuming slicing changes the original array

2. Which of the following is the correct syntax to get elements from index 2 up to but not including index 5 in Ruby array arr?

easy

A. arr[2...5]

B. arr[2:5]

C. arr[2..5]

D. arr[2-5]

Solution

1. Step 1: Identify the range operator for excluding end

   In Ruby, ... excludes the end index, so 2...5 means indexes 2, 3, and 4.

2. Step 2: Check syntax correctness

   Only arr[2...5] uses valid Ruby syntax for slicing excluding the end index.

3. Final Answer:

   arr[2...5] -> Option A

4. Quick Check:

   Use '...' to exclude end index in range [OK]

Hint: Use '...' to exclude the last index in range slicing [OK]

Common Mistakes:

• Using '..' when exclusion is needed
• Using Python-style slicing syntax like ':'
• Using invalid range syntax like '2-5'

3. What is the output of the following Ruby code?

arr = [5, 10, 15, 20, 25]
slice = arr[1...4]
puts slice.inspect

medium

A. [10, 15, 20, 25]

B. [15, 20, 25]

C. [10, 15, 20]

D. [5, 10, 15]

Solution

1. Step 1: Understand the slicing range

   The range 1...4 includes indexes 1, 2, and 3 but excludes 4.

2. Step 2: Extract elements at these indexes

   Elements at indexes 1, 2, 3 are 10, 15, and 20 respectively.

3. Final Answer:

   [10, 15, 20] -> Option C

4. Quick Check:

   Range '...' excludes end index = [10, 15, 20] [OK]

Hint: Remember '...' excludes the last index in slicing [OK]

Common Mistakes:

• Including element at index 4 mistakenly
• Confusing indexes with values
• Printing original array instead of slice

4. Identify the error in the following Ruby code snippet:

arr = [1, 2, 3, 4, 5]
slice = arr[2..6]
puts slice.inspect

medium

A. Syntax error due to invalid range

B. Range end index 6 is out of bounds but Ruby returns elements up to the end of the array

C. Runtime error because index 6 does not exist

D. slice will be empty array because range is invalid

Solution

1. Step 1: Check the range indexes against array size

   The array has indexes 0 to 4. The range 2..6 goes beyond the last index.

2. Step 2: Understand Ruby slicing behavior for out-of-bounds

   Ruby returns elements from index 2 to the end (indexes 2, 3, 4) and ignores indexes beyond 4 without error.

3. Final Answer:

   Range end index 6 is out of bounds but Ruby returns elements up to the end of the array -> Option B

4. Quick Check:

   Ruby slices safely even if range exceeds array size [OK]

Hint: Ruby slicing ignores out-of-bound indexes without error [OK]

Common Mistakes:

• Expecting runtime or syntax error
• Assuming slice will be empty
• Confusing Ruby with other languages that error here

5. Given arr = [2, 4, 6, 8, 10, 12], which expression returns a new array with every second element starting from index 1 up to index 5?

hard

A. arr[1..5].select.with_index { |_, i| i.even? }

B. arr[1..5].select.with_index { |_, i| i.odd? }

C. arr[1...6].values_at(*arr[1...6].each_index.select(&:even?))

D. arr[1..5].values_at(*arr[1..5].each_index.select(&:odd?))

Solution

1. Step 1: Slice array from index 1 to 5

   arr[1..5] gives [4, 6, 8, 10, 12].

2. Step 2: Select every second element starting from index 1 of the slice

   Indexes inside the slice are 0 to 4. Selecting odd indexes (1, 3) gives elements 6 and 10.

3. Step 3: Check options for correct syntax and output

   arr[1..5].values_at(*arr[1..5].each_index.select(&:odd?)) uses values_at with odd indexes on the slice correctly, returning [6, 10].

4. Final Answer:

   arr[1..5].values_at(*arr[1..5].each_index.select(&:odd?)) -> Option D

5. Quick Check:

   Use values_at with odd indexes on slice for every second element [OK]

Hint: Use values_at with selected indexes for custom slicing [OK]

Common Mistakes:

• Confusing odd/even indexes inside the slice
• Using select without values_at returns elements, not indexes
• Using wrong range operator causing off-by-one errors