Array Rotation Techniques in DSA Java - Time & Space Complexity

Time Complexity: Array Rotation Techniques

O(n)

Understanding Time Complexity

When we rotate an array, we move elements around to new positions. Understanding how long this takes helps us choose the best method.

We want to know how the time needed grows as the array gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public void rotateArray(int[] arr, int k) {
    int n = arr.length;
    k = k % n;
    int[] temp = new int[k];
    for (int i = 0; i < k; i++) {
        temp[i] = arr[n - k + i];
    }
    for (int i = n - 1; i >= k; i--) {
        arr[i] = arr[i - k];
    }
    for (int i = 0; i < k; i++) {
        arr[i] = temp[i];
    }
}
    

This code rotates the array to the right by k steps using a temporary array.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Three separate loops copying elements.
• How many times: Each loop runs roughly k or n times, where n is array size and k is rotation steps.

How Execution Grows With Input

As the array size grows, the number of operations grows mostly with n, since k is at most n.

Input Size (n)	Approx. Operations
10	About 30 (3 loops of 10 or less)
100	About 300
1000	About 3000

Pattern observation: Operations grow roughly in a straight line with n.

Final Time Complexity

Time Complexity: O(n)

This means the time to rotate grows directly with the size of the array.

Common Mistake

[X] Wrong: "Rotating by k steps always takes k operations only."

[OK] Correct: Because shifting elements involves moving many items, often close to n, not just k.

Interview Connect

Knowing how array rotation time grows helps you explain your choices clearly and shows you understand efficiency.

Self-Check

"What if we used the reversal method to rotate the array? How would the time complexity change?"

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