Array Rotation Techniques in DSA Go - Execution Trace

Concept Flow - Array Rotation Techniques

Start with array

↓

Choose rotation direction

↓

Rotate Left

↓

Shift elements

↓

Wrap elements around

↓

Resulting rotated array

This flow shows how an array is rotated either left or right by shifting elements and wrapping them around.

Execution Sample

DSA Go

arr := []int{1, 2, 3, 4, 5}
rotateLeft(arr, 2)
// Result: [3, 4, 5, 1, 2]

Rotate the array left by 2 positions, moving first two elements to the end.

Execution Table

Step	Operation	Array State	Pointer/Index Changes	Visual State
1	Start with original array	[1, 2, 3, 4, 5]	head=0	1 -> 2 -> 3 -> 4 -> 5 -> null
2	Rotate left by 1: shift elements left	[2, 3, 4, 5, 5]	shift indexes left	2 -> 3 -> 4 -> 5 -> 5 -> null
3	Wrap first element to end	[2, 3, 4, 5, 1]	moved 1 to end	2 -> 3 -> 4 -> 5 -> 1 -> null
4	Rotate left by 2: repeat shift	[3, 4, 5, 1, 1]	shift indexes left	3 -> 4 -> 5 -> 1 -> 1 -> null
5	Wrap next element to end	[3, 4, 5, 1, 2]	moved 2 to end	3 -> 4 -> 5 -> 1 -> 2 -> null
6	Rotation complete	[3, 4, 5, 1, 2]	final state	3 -> 4 -> 5 -> 1 -> 2 -> null

💡 Rotation stops after shifting and wrapping elements 2 times as requested.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	Final
arr	[1, 2, 3, 4, 5]	[2, 3, 4, 5, 5]	[2, 3, 4, 5, 1]	[3, 4, 5, 1, 1]	[3, 4, 5, 1, 2]	[3, 4, 5, 1, 2]
head	0	0	0	0	0	0
rotation_count	0	1	1	2	2	2

Key Moments - 3 Insights

Why do we shift elements left before wrapping the first element to the end?

Because shifting left moves all elements one position forward, freeing the last position to place the wrapped element. See steps 2 and 3 in the execution_table.

Why does the array temporarily have duplicate elements during shifting?

During shifting, the last element is overwritten temporarily until we place the wrapped element there. This is shown in steps 2 and 4 where the last element duplicates before wrapping.

What happens if rotation count is zero or equal to array length?

No rotation occurs because shifting zero times or full length results in the same array. The exit_note explains stopping after requested rotations.

Visual Quiz - 3 Questions

Look at the execution_table, what is the array state after step 3?

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

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

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

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

💡 Hint

Check the 'Array State' column for step 3 in the execution_table.

At which step does the rotation count reach 2 according to variable_tracker?

A After Step 2

B After Step 3

C After Step 4

D After Step 5

💡 Hint

Look at the 'rotation_count' row in variable_tracker to see when it becomes 2.

If we rotate left by 1 instead of 2, what would be the final array state?

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

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

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

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

💡 Hint

Refer to the array state after step 3 which shows rotation by 1.

Concept Snapshot

Array Rotation Techniques:
- Rotate left: shift elements left, wrap first elements to end.
- Rotate right: shift elements right, wrap last elements to front.
- Rotation count defines how many times to shift and wrap.
- Temporary duplicates occur during shifting.
- Final array shows rotated order after all shifts.

Full Transcript

This visual execution traces rotating an array left by 2 positions. Starting with [1, 2, 3, 4, 5], each rotation shifts elements left by one and moves the first element to the end. After two rotations, the array becomes [3, 4, 5, 1, 2]. The execution table shows each step's array state and pointer changes. The variable tracker monitors the array and rotation count. Key moments clarify why shifting happens before wrapping and why duplicates appear temporarily. The quiz tests understanding of array states and rotation counts at specific steps. The snapshot summarizes the rotation process and key rules.

Practice

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

easy

A. Reverses the entire array

B. Moves the last element to the front and shifts others right by one

C. Moves the first element to the end and shifts others left by one

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.

2. Step 2: Effect on array elements

   All other elements shift one position to the left.

3. Final Answer:

   Moves the first element to the end and shifts others left by one -> Option C

4. Quick Check:

   Left rotate 1 = first element to end [OK]

Hint: Left rotate by 1 moves front element to back [OK]

Common Mistakes:

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

2. Which of the following Go code snippets correctly rotates an array arr to the right by 1 position?

easy

A. arr = append(arr[len(arr)-1:], arr[:len(arr)-1])

B. arr = append(arr[1:], arr[0])

C. arr = append(arr[:len(arr)-1], arr[len(arr)-1])

D. arr = append([]int{arr[len(arr)-1]}, arr[:len(arr)-1]...)

Solution

1. Step 1: Understand right rotation by 1

   Right rotation by 1 moves last element to front.

2. Step 2: Check code correctness

   arr = append([]int{arr[len(arr)-1]}, arr[:len(arr)-1]...) prepends last element to the rest correctly using slicing and append.

3. Final Answer:

   arr = append([]int{arr[len(arr)-1]}, arr[:len(arr)-1]...) -> Option D

4. Quick Check:

   Right rotate 1 = last element front [OK]

Hint: Right rotate by 1 prepends last element [OK]

Common Mistakes:

• Using wrong slice indices
• Appending single element without slice expansion
• Confusing left and right rotation code

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

arr := []int{10, 20, 30, 40, 50}
k := 2
n := len(arr)
k = k % n
rotated := append(arr[n-k:], arr[:n-k]...)
fmt.Println(rotated)

medium

A. [40 50 10 20 30]

B. [30 40 50 10 20]

C. [20 30 40 50 10]

D. [50 10 20 30 40]

Solution

1. Step 1: Calculate effective rotation

   k=2, n=5, so k%5=2 means rotate right by 2.

2. Step 2: Perform rotation

   arr[n-k:] = arr[3:] = [40,50], arr[:n-k] = arr[:3] = [10,20,30]. Append gives [40,50,10,20,30].

3. Final Answer:

   [40 50 10 20 30] -> Option A

4. Quick Check:

   Right rotate 2 = last 2 front [OK]

Hint: Right rotate by k = last k elements front [OK]

Common Mistakes:

• Not using modulo for k
• Swapping slice indices incorrectly
• Confusing left and right rotation

4. Identify the error in this Go code that attempts to left rotate an array by 3 positions:

arr := []int{1, 2, 3, 4, 5}
k := 3
rotated := append(arr[k:], arr[k:]...)
fmt.Println(rotated)

medium

A. The slice is appended twice from the same index, missing first part

B. The variable k is not used correctly for rotation

C. The array length is not considered for rotation

D. The append function is used incorrectly without slice expansion

Solution

1. Step 1: Analyze append arguments

   Both slices are arr[k:], so elements from index 3 to end repeated twice.

2. Step 2: Identify missing part

   Left rotation requires appending arr[:k] after arr[k:], but code repeats arr[k:] instead.

3. Final Answer:

   The slice is appended twice from the same index, missing first part -> Option A

4. Quick Check:

   Left rotate needs arr[k:] + arr[:k] [OK]

Hint: Append arr[k:] then arr[:k] for left rotation [OK]

Common Mistakes:

• Appending same slice twice
• Ignoring first k elements in rotation
• Not using slice expansion operator

5. Given an array arr = []int{5, 10, 15, 20, 25, 30}, how would you rotate it left by 4 positions using Go slices? Choose the correct code snippet.

hard

A. rotated := append(arr[:4], arr[4:]...)

B. rotated := append(arr[4:], arr[:4]...)

C. rotated := append(arr[len(arr)-4:], arr[:len(arr)-4]...)

D. rotated := append(arr[len(arr)-4:], arr[len(arr)-4:]...)

Solution

1. Step 1: Understand left rotation by 4

   Left rotation by 4 moves first 4 elements to the end.

2. Step 2: Use slicing correctly

   arr[4:] gives elements from index 4 to end, arr[:4] gives first 4 elements. Append arr[4:] + arr[:4] rotates left by 4.

3. Final Answer:

   rotated := append(arr[4:], arr[:4]...) -> Option B

4. Quick Check:

   Left rotate 4 = arr[4:] + arr[:4] [OK]

Hint: Left rotate k = arr[k:] + arr[:k] [OK]

Common Mistakes:

• Swapping slice order
• Using wrong slice indices
• Confusing left and right rotation