Array Rotation Techniques in DSA Rust - Time & Space Complexity

Time Complexity: Array Rotation Techniques

O(n)

Understanding Time Complexity

When we rotate an array, we move its elements around. Understanding how long this takes helps us choose the best way to do it.

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.

fn rotate_array(arr: &mut [i32], k: usize) {
    let n = arr.len();
    let k = k % n;
    arr.reverse();
    arr[0..k].reverse();
    arr[k..n].reverse();
}
    

This code rotates the array to the right by k steps using three reversals.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Reversing parts of the array by swapping elements.
• How many times: Each reversal touches elements once, total of n elements reversed in three parts.

How Execution Grows With Input

As the array size grows, the number of swaps grows roughly the same way.

Input Size (n)	Approx. Operations
10	About 10 swaps
100	About 100 swaps
1000	About 1000 swaps

Pattern observation: The work grows linearly with the size of the array.

Final Time Complexity

Time Complexity: O(n)

This means the time to rotate grows directly with the number of elements in the array.

Common Mistake

[X] Wrong: "Rotating by k steps takes k times longer than rotating by 1 step."

[OK] Correct: This method touches each element a fixed number of times regardless of k, so time depends on n, not k.

Interview Connect

Knowing how array rotation works and its time cost helps you explain and choose efficient solutions in real coding tasks.

Self-Check

"What if we used a temporary array to store rotated elements? How would the time complexity change?"

Practice

1. What does rotating an array to the left by 2 positions mean for the array [1, 2, 3, 4, 5]?

easy

A. The last two elements move to the front, resulting in [4, 5, 1, 2, 3]

B. The first two elements move to the end, resulting in [3, 4, 5, 1, 2]

C. The array is reversed, resulting in [5, 4, 3, 2, 1]

D. The array remains unchanged

Solution

1. Step 1: Understand left rotation by 2

   Left rotation by 2 means removing the first two elements and placing them at the end.

2. Step 2: Apply rotation on given array

   Starting with [1, 2, 3, 4, 5], removing first two elements [1, 2] and appending them at the end gives [3, 4, 5, 1, 2].

3. Final Answer:

   [3, 4, 5, 1, 2] -> Option B

4. Quick Check:

   Left rotate by 2 = first 2 elements moved to end [OK]

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

Common Mistakes:

• Confusing left rotation with right rotation
• Not moving elements but just swapping
• Thinking rotation reverses the array

2. Which of the following Rust code snippets correctly rotates an array arr of length 5 to the right by 1 position?

easy

A. let rotated = [&arr[0..4], &arr[4..5]].concat();

B. let rotated = [&arr[1..5], &arr[0..1]].concat();

C. let rotated = [&arr[4..5], &arr[0..4]].concat();

D. let rotated = [&arr[3..5], &arr[0..3]].concat();

Solution

1. Step 1: Understand right rotation by 1

   Right rotation by 1 means last element moves to front, rest shift right.

2. Step 2: Check code slices for correct order

   let rotated = [&arr[4..5], &arr[0..4]].concat(); takes last element arr[4..5] and then first 4 elements arr[0..4], concatenating them correctly.

3. Final Answer:

   let rotated = [&arr[4..5], &arr[0..4]].concat(); -> Option C

4. Quick Check:

   Right rotate by 1 = last element + rest [OK]

Hint: Right rotate by 1 = last element + rest [OK]

Common Mistakes:

• Using wrong slice indices
• Concatenating slices in wrong order
• Confusing left and right rotation slices

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

let arr = vec![10, 20, 30, 40, 50];
let k = 2;
let n = arr.len();
let rotated: Vec<_> = arr.iter().cycle().skip(n - k).take(n).cloned().collect();
println!("{:?}", rotated);

medium

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

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

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

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

Solution

1. Step 1: Understand rotation logic in code

   Code cycles array, skips n-k=3 elements, then takes n=5 elements, effectively right rotating by 2.

2. Step 2: Trace elements after skipping 3

   Skipping first 3 elements: 10,20,30 skipped; next elements are 40,50, then cycle back 10,20,30.

3. Step 3: Collect 5 elements

   Elements collected: 40, 50, 10, 20, 30.

4. Final Answer:

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

5. Quick Check:

   Right rotate by 2 = last 2 elements front [OK]

Hint: Cycle and skip n-k for right rotation [OK]

Common Mistakes:

• Confusing skip count with k
• Mixing left and right rotation logic
• Not using cycle iterator correctly

4. The following Rust function is intended to left rotate an array by k positions. What is the bug?

fn left_rotate(arr: &mut [i32], k: usize) {
    let n = arr.len();
    for i in 0..k {
        let temp = arr[i];
        arr[i] = arr[n - k + i];
        arr[n - k + i] = temp;
    }
}

medium

A. It swaps wrong elements; rotation logic is incorrect

B. It causes index out of bounds error

C. It reverses the array instead of rotating

D. It does nothing because loop range is zero

Solution

1. Step 1: Analyze the swapping logic

   The code swaps arr[i] with arr[n-k+i], which does not achieve left rotation but swaps wrong pairs.

2. Step 2: Understand correct rotation method

   Left rotation requires shifting elements left and moving first k elements to end, not swapping like this.

3. Final Answer:

   It swaps wrong elements; rotation logic is incorrect -> Option A

4. Quick Check:

   Incorrect swap logic = wrong rotation [OK]

Hint: Rotation needs shift + append, not simple swaps [OK]

Common Mistakes:

• Swapping instead of shifting elements
• Using wrong indices for swapping
• Assuming swap equals rotation

5. Given an array arr = [1, 2, 3, 4, 5, 6], how can you rotate it left by 4 positions using Rust slice operations without extra space?

hard

A. Reverse first 4 elements, reverse last 2 elements, reverse entire array

B. Reverse last 4 elements, reverse first 2 elements, reverse entire array

C. Reverse entire array, reverse last 2 elements, reverse first 4 elements

D. Reverse first 2 elements, reverse last 4 elements, reverse entire array

Solution

1. Step 1: Understand rotation by reversal method

   Left rotate by k means reverse first k elements, reverse remaining, then reverse whole array.

2. Step 2: Apply to k=4, n=6

   Reverse first 4 elements [1,2,3,4], reverse last 2 elements [5,6], then reverse entire array.

3. Final Answer:

   Reverse first 4 elements, reverse last 2 elements, reverse entire array -> Option A

4. Quick Check:

   Rotation by reversal = reverse k, reverse n-k, reverse all [OK]

Hint: Rotate left k = reverse first k, reverse rest, reverse all [OK]

Common Mistakes:

• Reversing in wrong order
• Mixing k and n-k segments
• Using extra space instead of in-place reversal