Array Rotation Techniques in DSA Typescript - 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 changes as the array grows larger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function rotateArray(arr: number[], k: number): number[] {
  const n = arr.length;
  k = k % n;
  const rotated = arr.slice(n - k).concat(arr.slice(0, n - k));
  return rotated;
}
    

This code rotates the array to the right by k steps using slicing and concatenation.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Creating two slices of the array and then joining them.
• How many times: Each slice goes through part of the array once, so together they cover all elements once.

How Execution Grows With Input

As the array size grows, the number of elements we copy grows too.

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

Pattern observation: The work grows directly with the number of elements, so doubling the array size roughly doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to rotate grows in a straight line with the array size.

Common Mistake

[X] Wrong: "Rotating an array by k steps always takes constant time because k is fixed."

[OK] Correct: Even if k is fixed, the code copies parts of the array proportional to its size, so time grows with the array length.

Interview Connect

Knowing how array rotation works and its time cost helps you explain your choices clearly and shows you understand how data moves in memory.

Self-Check

"What if we rotated the array in place without creating a new array? How would the time complexity change?"

Practice

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

easy

A. [4, 1, 2, 3]

B. [2, 3, 4, 1]

C. [1, 2, 3, 4]

D. [3, 4, 1, 2]

Solution

1. Step 1: Understand right rotation

   Right rotation by 1 means the last element moves to the front, shifting others right.

2. Step 2: Apply to example array

   Last element 4 moves to front, rest shift right: [4, 1, 2, 3].

3. Final Answer:

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

4. Quick Check:

   Right rotate by 1 = last element front [OK]

Hint: Right rotate by 1 means last element moves front [OK]

Common Mistakes:

• Confusing right rotation with left rotation
• Not moving the last element to front
• Shifting elements incorrectly

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

easy

A. arr.push(arr.shift()!)

B. arr.unshift(arr.pop()!)

C. arr.pop(); arr.unshift(arr.pop()!)

D. arr.shift(); arr.push(arr.shift()!)

Solution

1. Step 1: Understand left rotation by 1

   Left rotation by 1 means first element moves to the end.

2. Step 2: Check code snippets

   arr.push(arr.shift()!) uses arr.shift() to remove first element and arr.push() to add it at end, correct for left rotation.

3. Final Answer:

   arr.push(arr.shift()!) -> Option A

4. Quick Check:

   Left rotate = shift then push [OK]

Hint: Left rotate by 1 = remove first, add to end [OK]

Common Mistakes:

• Using pop instead of shift for left rotation
• Using unshift with pop for left rotation
• Calling shift or pop twice incorrectly

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

const arr = [10, 20, 30, 40, 50];
const k = 2;
const rotated = arr.slice(-k).concat(arr.slice(0, arr.length - k));
console.log(rotated);

medium

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

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

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

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

Solution

1. Step 1: Understand slicing for rotation

   arr.slice(-k) takes last 2 elements: [40, 50]. arr.slice(0, arr.length - k) takes first 3 elements: [10, 20, 30].

2. Step 2: Concatenate slices

   Concatenating gives [40, 50, 10, 20, 30], which is right rotation by 2.

3. Final Answer:

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

4. Quick Check:

   Right rotate by 2 = last 2 front [OK]

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

Common Mistakes:

• Using slice with wrong indices
• Confusing left and right rotation
• Not concatenating slices in correct order

4. Find the error in this TypeScript function that rotates an array to the left by k positions:

function rotateLeft(arr: number[], k: number): number[] {
  for (let i = 0; i < k; i++) {
    arr.push(arr.shift());
  }
  return arr;
}

medium

A. arr.push() should be arr.unshift()

B. Loop should run from i = 1 to k

C. arr.shift() can return undefined, causing push(undefined)

D. Function should return void, not number[]

Solution

1. Step 1: Check arr.shift() return type

   arr.shift() returns element or undefined if array empty. Here, no check for undefined.

2. Step 2: Effect on arr.push()

   If arr.shift() returns undefined, arr.push(undefined) adds undefined to array, causing incorrect result or runtime issues.

3. Final Answer:

   arr.shift() can return undefined, causing push(undefined) -> Option C

4. Quick Check:

   shift may be undefined, must check [OK]

Hint: Check shift() return before push to avoid undefined [OK]

Common Mistakes:

• Ignoring undefined from shift()
• Wrong loop range assumptions
• Confusing push and unshift methods

5. Given an array arr of length n, how can you rotate it to the left by k positions efficiently without using extra space? Choose the correct approach.

hard

A. Use a temporary array to store first k elements, then shift rest left, then copy temp at end

B. Sort the array and then rotate by k positions

C. Rotate one by one k times using shift and push

D. Reverse first k elements, reverse last n-k elements, then reverse whole array

Solution

1. Step 1: Understand in-place rotation using reversal

   Reversing parts of array and then whole array rotates left by k without extra space.

2. Step 2: Apply reversal steps

   Reverse first k elements, reverse last n-k elements, then reverse entire array to get rotated array.

3. Final Answer:

   Reverse first k, reverse last n-k, reverse whole array -> Option D

4. Quick Check:

   Reversal method = in-place left rotate [OK]

Hint: Use triple reversal for in-place left rotation [OK]

Common Mistakes:

• Using extra array when not needed
• Rotating one by one causing O(n*k) time
• Sorting array unrelated to rotation