Array Rotation Techniques in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to rotate the array to the right by one position.

DSA Java

int last = arr[arr.length [1] 1];
for (int i = arr.length - 1; i > 0; i--) {
    arr[i] = arr[i - 1];
}
arr[0] = last;

Drag options to blanks, or click blank then click option'

A*

B+

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using + instead of - causes ArrayIndexOutOfBoundsException.

Using * or / is incorrect for indexing.

We subtract 1 from arr.length to get the last index of the array.

2fill in blank

medium

Complete the code to rotate the array to the left by one position.

DSA Java

int first = arr[0];
for (int i = 0; i < arr.length - 1; i++) {
    arr[i] = arr[i [1] 1];
}
arr[arr.length - 1] = first;

Drag options to blanks, or click blank then click option'

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using - instead of + causes wrong indexing.

Using * or / is invalid for array indices.

We add 1 to i to get the next element for shifting left.

3fill in blank

hard

Fix the error in the code to rotate the array to the right by k positions.

DSA Java

k = k % arr.length;
reverse(arr, 0, arr.length - 1);
reverse(arr, 0, k [1] 1);
reverse(arr, k, arr.length - 1);

Drag options to blanks, or click blank then click option'

A/

B+

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using + causes index out of bounds.

Using * or / is invalid for indexing.

We subtract 1 from k to get the correct end index for the first reverse.

4fill in blank

hard

Fill both blanks to create a method that reverses a part of the array from start to end indices.

DSA Java

void reverse(int[] arr, int start, int end) {
    while (start [1] end) {
        int temp = arr[start];
        arr[start] = arr[end];
        arr[end] = temp;
        start [2] 1;
        end -= 1;
    }
}

Drag options to blanks, or click blank then click option'

A<

B>

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < causes the loop to never run.

Decrementing start instead of incrementing causes infinite loop.

We continue while start is less than end, and increment start by 1 each loop.

5fill in blank

hard

Fill all three blanks to create a method that rotates the array to the left by d positions using reversal.

DSA Java

void leftRotate(int[] arr, int d) {
    int n = arr.length;
    d = d % n;
    reverse(arr, 0, d [1] 1);
    reverse(arr, d, n [2] 1);
    reverse(arr, 0, n [3] 1);
}

Drag options to blanks, or click blank then click option'

A-

B+

Attempts:

3 left

💡 Hint

Common Mistakes

Using + instead of - causes index out of bounds errors.

Not adjusting indices leads to incorrect rotation.

We subtract 1 from indices to get correct zero-based end indices for reversal.

Practice

1. What does rotating an array to the left by 1 position mean?
Example: [1, 2, 3, 4] rotated left by 1 becomes?

easy

A. [2, 3, 4, 1]

B. [4, 1, 2, 3]

C. [1, 4, 3, 2]

D. [3, 4, 1, 2]

Solution

1. Step 1: Understand left rotation

   Left rotation by 1 means each element moves one step to the left, and the first element moves to the end.

2. Step 2: Apply left rotation on example

   For [1, 2, 3, 4], after left rotation by 1, it becomes [2, 3, 4, 1].

3. Final Answer:

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

4. Quick Check:

   Left rotate by 1 moves first element to end [OK]

Hint: Left rotate shifts elements left, first goes to last [OK]

Common Mistakes:

• Confusing left rotation with right rotation
• Not moving the first element to the end
• Shifting elements in the wrong direction

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

easy

A. int last = arr[n-1]; for(int i = n-1; i > 0; i--) arr[i] = arr[i-1]; arr[0] = last;

B. int first = arr[0]; for(int i = 0; i < n-1; i++) arr[i] = arr[i+1]; arr[n-1] = first;

C. for(int i = 0; i < n; i++) arr[i] = arr[(i+1) % n];

D. int temp = arr[0]; arr[0] = arr[n-1]; arr[n-1] = temp;

Solution

1. Step 1: Understand right rotation by 1

   Right rotation by 1 moves each element one step right, last element moves to first position.

2. Step 2: Check code correctness

   int last = arr[n-1]; for(int i = n-1; i > 0; i--) arr[i] = arr[i-1]; arr[0] = last; saves last element, shifts elements right, then sets first element to saved last. This matches right rotation.

3. Final Answer:

   int last = arr[n-1]; for(int i = n-1; i > 0; i--) arr[i] = arr[i-1]; arr[0] = last; -> Option A

4. Quick Check:

   Right rotate shifts right, last to first [OK]

Hint: Right rotate moves last element to front [OK]

Common Mistakes:

• Using left rotation logic for right rotation
• Incorrect loop direction causing overwrite
• Swapping only first and last elements

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

int[] arr = {10, 20, 30, 40, 50};
int d = 2;
for(int i = 0; i < d; i++) {
    int temp = arr[0];
    for(int j = 0; j < arr.length - 1; j++) {
        arr[j] = arr[j+1];
    }
    arr[arr.length - 1] = temp;
}
System.out.println(java.util.Arrays.toString(arr));

medium

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

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

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

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

Solution

1. Step 1: Understand the loop rotates array left by 1, repeated d times

   Each iteration moves first element to end, shifting left by 1. Loop runs twice (d=2).

2. Step 2: Trace array after each rotation

   After 1st rotation: [20, 30, 40, 50, 10]
   After 2nd rotation: [30, 40, 50, 10, 20]

3. Final Answer:

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

4. Quick Check:

   Left rotate by 2 shifts first 2 elements to end [OK]

Hint: Left rotate by d moves first d elements to end [OK]

Common Mistakes:

• Confusing left and right rotation
• Not repeating rotation d times
• Off-by-one errors in loops

4. Identify the error in this Java code that attempts to rotate an array arr of length n to the left by 1:

int temp = arr[0];
for(int i = 0; i < n; i++) {
    arr[i] = arr[i+1];
}
arr[n-1] = temp;

medium

A. temp should be assigned after the loop

B. Array length n is incorrect

C. Loop runs out of bounds accessing arr[n]

D. No error, code is correct

Solution

1. Step 1: Analyze the loop range and array access

   The loop runs from i=0 to i=n-1, but inside it accesses arr[i+1]. For i=n-1, arr[i+1] = arr[n], which is out of bounds.

2. Step 2: Identify the cause of error

   Accessing arr[n] causes ArrayIndexOutOfBoundsException. The loop should run only till i < n-1.

3. Final Answer:

   Loop runs out of bounds accessing arr[n] -> Option C

4. Quick Check:

   Array index must be within 0 to n-1 [OK]

Hint: Check array indices carefully to avoid out-of-bounds [OK]

Common Mistakes:

• Looping till i < n instead of i < n-1
• Ignoring array length limits
• Misplacing temp assignment

5. Given an array arr = [5, 10, 15, 20, 25, 30], how can you rotate it to the right by 3 positions efficiently without using extra space?

hard

A. Swap elements pairwise from start to end 3 times

B. Shift elements one by one to the right 3 times using nested loops

C. Copy last 3 elements to a temp array, shift rest right, then copy temp to front

D. Reverse entire array, reverse first 3 elements, then reverse remaining elements

Solution

1. Step 1: Understand the reversal algorithm for rotation

   To rotate right by d, reverse whole array, then reverse first d elements, then reverse remaining n-d elements.

2. Step 2: Apply to example

   Array: [5,10,15,20,25,30]
   Reverse all: [30,25,20,15,10,5]
   Reverse first 3: [20,25,30,15,10,5]
   Reverse last 3: [20,25,30,5,10,15]

3. Final Answer:

   Reverse entire array, reverse first 3 elements, then reverse remaining elements -> Option D

4. Quick Check:

   Reversal method rotates array efficiently without extra space [OK]

Hint: Use reversal method: reverse all, then parts [OK]

Common Mistakes:

• Using extra space instead of in-place reversal
• Shifting elements one by one inefficiently
• Incorrect order of reversals