Array Rotation Techniques in DSA Python - Build from Scratch

Array Rotation Techniques

📖 Scenario: Imagine you have a list of daily tasks arranged in order. Sometimes, you want to start your day from a different task but keep the order of the remaining tasks the same. This is like rotating the list of tasks.

🎯 Goal: You will create a program that rotates an array (list) to the left by a given number of positions. This means the first few tasks move to the end, and the rest shift forward.

📋 What You'll Learn

Create a list called tasks with exact values: ['Email', 'Meeting', 'Coding', 'Lunch', 'Review']

Create a variable called rotate_by and set it to 2

Write code to rotate the tasks list to the left by rotate_by positions

Print the rotated list exactly as a Python list

💡 Why This Matters

🌍 Real World

Rotating lists is useful in scheduling, games, and data processing where order changes but elements stay the same.

💼 Career

Understanding array rotation helps in coding interviews and tasks involving data manipulation and optimization.

1

Create the initial list of tasks

Create a list called tasks with these exact values in order: 'Email', 'Meeting', 'Coding', 'Lunch', 'Review'

DSA Python

# Create the list called tasks with the given values
# Your code here

Hint

Use square brackets [] to create a list and separate items with commas.

2

Set the number of positions to rotate

Create a variable called rotate_by and set it to the number 2

DSA Python

tasks = ['Email', 'Meeting', 'Coding', 'Lunch', 'Review']
# Create rotate_by variable and set it to 2
# Your code here

Hint

Just assign the number 2 to the variable rotate_by.

3

Rotate the list to the left by rotate_by positions

Create a new list called rotated_tasks that contains the tasks list rotated to the left by rotate_by positions using slicing

DSA Python

tasks = ['Email', 'Meeting', 'Coding', 'Lunch', 'Review']
rotate_by = 2
# Create rotated_tasks by slicing tasks to rotate left by rotate_by
# Your code here

Hint

Use list slicing: tasks[rotate_by:] gets the part after the first rotate_by items, and tasks[:rotate_by] gets the first rotate_by items.

4

Print the rotated list

Print the rotated_tasks list exactly as a Python list

DSA Python

tasks = ['Email', 'Meeting', 'Coding', 'Lunch', 'Review']
rotate_by = 2
rotated_tasks = tasks[rotate_by:] + tasks[:rotate_by]
# Print rotated_tasks
# Your code here

Hint

Use print(rotated_tasks) to show the rotated list.

Practice

1. What does rotating an array to the left by 1 position do to the elements?

easy

A. Moves each element one step to the right, last element goes to the front

B. Moves each element one step to the left, first element goes to the end

C. Reverses the array

D. Sorts the array in ascending order

Solution

1. Step 1: Understand left rotation

   Left rotation by 1 means the element at index 0 moves to the end, and all others shift left by one.

2. Step 2: Visualize the movement

   For example, [1, 2, 3] becomes [2, 3, 1] after left rotation by 1.

3. Final Answer:

   Moves each element one step to the left, first element goes to the end -> Option B

4. Quick Check:

   Left rotation = shift left + wrap first element [OK]

Hint: Left rotation shifts left, first element moves to end [OK]

Common Mistakes:

• Confusing left rotation with right rotation
• Thinking rotation sorts the array
• Assuming rotation reverses the array

2. Which of the following is the correct Python syntax to rotate an array arr to the right by 2 positions using slicing?

easy

A. arr = arr[-2:] + arr[:-2]

B. arr = arr[2:] + arr[:2]

C. arr = arr[:2] + arr[2:]

D. arr = arr[-2:] - arr[:-2]

Solution

1. Step 1: Understand right rotation with slicing

   Right rotation by 2 means last 2 elements move to front, rest shift right.

2. Step 2: Apply slicing syntax

   arr[-2:] gets last 2 elements, arr[:-2] gets all except last 2. Concatenating gives rotated array.

3. Final Answer:

   arr = arr[-2:] + arr[:-2] -> Option A

4. Quick Check:

   Right rotation by 2 = last 2 + rest [OK]

Hint: Right rotation by k = last k elements + rest [OK]

Common Mistakes:

• Using + operator incorrectly with slices
• Confusing left and right rotation slices
• Using subtraction operator instead of concatenation

3. What is the output of the following code?

arr = [10, 20, 30, 40, 50]
k = 3
rotated = arr[k:] + arr[:k]
print(rotated)

medium

A. [50, 10, 20, 30, 40]

B. [30, 40, 50, 10, 20]

C. [20, 30, 40, 50, 10]

D. [40, 50, 10, 20, 30]

Solution

1. Step 1: Identify rotation type

   arr[k:] + arr[:k] rotates array left by k=3 positions.

2. Step 2: Perform slicing and concatenation

   arr[3:] = [40, 50], arr[:3] = [10, 20, 30], combined = [40, 50, 10, 20, 30]

3. Final Answer:

   [40, 50, 10, 20, 30] -> Option D

4. Quick Check:

   Left rotation by 3 = [40,50] + [10,20,30] [OK]

Hint: arr[k:] + arr[:k] = left rotate by k [OK]

Common Mistakes:

• Mixing left and right rotation logic
• Incorrect slicing indices
• Misreading k value

4. The following code is intended to rotate an array arr to the right by k positions. What is the error?

def rotate_right(arr, k):
    n = len(arr)
    k = k % n
    rotated = arr[k:] + arr[:k]
    return rotated

print(rotate_right([1,2,3,4,5], 2))

medium

A. The slicing indices perform left rotation, not right rotation

B. The modulo operation is incorrect

C. The function does not return anything

D. The print statement is missing parentheses

Solution

1. Step 1: Analyze slicing for rotation

   arr[k:] + arr[:k] rotates array left by k, not right.

2. Step 2: Correct right rotation slicing

   Right rotation by k should be arr[-k:] + arr[:-k].

3. Final Answer:

   The slicing indices perform left rotation, not right rotation -> Option A

4. Quick Check:

   Right rotation needs arr[-k:] + arr[:-k] [OK]

Hint: Right rotation uses negative slicing indices [OK]

Common Mistakes:

• Using left rotation slices for right rotation
• Ignoring modulo to handle k > n
• Forgetting to return the rotated array

5. Given an array arr = [3, 8, 9, 7, 6], how would you rotate it to the right by k = 3 positions using a single Python statement without slicing?

hard

A. Use arr.sort() then reverse

B. Use arr.reverse() three times

C. Use a loop to pop last element and insert at front 3 times

D. Use arr.append(arr.pop()) once

Solution

1. Step 1: Understand rotation without slicing

   Rotating right by 3 means moving last element to front 3 times.

2. Step 2: Use loop with pop and insert

   Loop 3 times: pop last element and insert at index 0 to rotate right.

3. Final Answer:

   Use a loop to pop last element and insert at front 3 times -> Option C

4. Quick Check:

   Pop + insert at front repeated k times = right rotation [OK]

Hint: Pop last and insert front k times for right rotation [OK]

Common Mistakes:

• Using reverse instead of rotation
• Sorting array mistakenly
• Only popping once instead of k times